Class 11 Commerce Maths - Straight Lines

Find the slope of the line passing through the points $A(2, 3)$ and $B(4, 7)$ .

Fact: Slope of line passing through the points

$(x_1,y_1)$ and

$(x_2,y_2)$ is given by,

$m = \dfrac{y_2-y_1}{x_2-x_1}$

Therefore slope of line passing through the points

$A(2,3)$ and

$B(4,7)$ is

$=\dfrac{7-3}{4-2}=\dfrac{4}{2}=2$

Find the slope of line having inclination ${60}^{o}$ .

Inclination of the line is

${60}^{o}$ .

$\therefore \theta = {60}^{o}$
Now, the slope of the line

$= \tan{\theta} = \tan{{60}^{o}}=\sqrt{3}$ .

Find the slope of the line passing through the points $C(3, 5)$ and $D(-2, -3)$

$C \equiv (3, 5) = (x_{1}, y_{1})$
and

$D \equiv (-2, -3) = (x_{2}, y_{2})$
Then, Slope of line

$CD = \dfrac {y_{1} - y_{2}}{x_{1} - x_{2}}$

$= \dfrac {5 - (-3)}{3 - (-2)}$

$= \dfrac {5 + 3}{3 + 2} = \dfrac {8}{5}$

If the straight line joining two points $P(5,8)$ and $Q (8,k)$ is parallel to $x-axis$ , then write the value of $k$ .

As we know that slope

$\left( m \right)$ of the line joining the two points

$\left( {x}_{1}, {y}_{1} \right)$ and

$\left( {x}_{2}, {y}_{2} \right)$ , is given as-

$m = \cfrac{\left( {y}_{2} - {y}_{1} \right)}{\left( {x}_{2} - {x}_{1} \right)}$

Therefore, slope of line joining the points

$\left( 5, 8 \right)$ and

$\left( 8, k \right)$ is-

$m = \cfrac{k - 8}{8 - 5} = \cfrac{\left( k - 8 \right)}{3}$

Now,

Slope of

$x$ -axis

$= 0$

As we know that if lines are parallel then they have the same slope.

$\therefore \cfrac{\left( k - 8 \right)}{3} = 0$

$\Rightarrow k - 8 = 0$

$\Rightarrow k = 8$

Thus the value of

$k$ is

$8$ .

Hence the correct answer is

$8$ .

If the distance between the points $(4,\,p)$ and $(1,\,0)$ is $5$ , then the find the value of $p$ .

1338028_1275888_ans_b209d11891cf4500abd6496a49981aed.PNG

The points $(-2,5)$ and $(3,-5)$ are plotted in xy planes. Find the slope and y intercept of the line joining the points.

Slope(m)

$= \frac{{5 + 5}}{{ - 2 - 3}} = - 2$

$\begin{array}{l} y-5=-2\left( { x+2 } \right) \\ y-5=-2x-4 \\ y+2x=1 \end{array}$

$(0,1)$ Slope

$=-2$

$y$ intercept

$=1[(0,1)]$

1212536_1305957_ans_6e27ee6437064096bba1292f22c3a87c.PNG

Find the distance between $\left(-4,5\right)$ from the origin.

Distance between

$\left(-4,5\right)$ from the origin

$=$ Distance between

$\left(-4,5\right)$ from

$\left(0,0\right)$

$=\sqrt{ { \left( {x_2}-{ { x_1 } } \right) }^{ 2}+{ \left( {y_2 }-{ y_1 } \right) }^{ 2 } }$

$=\sqrt{{\left(-4-0\right)}^{2}+{\left(5-0\right)}^{2}}$

$=\sqrt{16+25}$

$=\sqrt{41}$ units.

Solve the following question:
Find the slope of the line passing through the points $A(2, 3)$ and $B(4, 7)$ .

Given

$A(2,3)$ and

$B(4,7)$

Slope of the line passing through

$(x_1,y_2)$ and

$(x_2,y_2)$ is

$m=\dfrac {y_{2} - y_{1}}{x_{2} - x_{1}}$

Here

$(x_1,y_2)=A(2,3)$ and

$(x_2,y_2)=(4,7)$

$m=\dfrac {7-3}{4-2}=\dfrac 42=2$

$\therefore$ the slope of the line passing through the points

$A(2,3)$ and

$B(4,7)$ is

$2$

The type of triangle formed by the points A (5, 6), B (4, 2) and C (7, 5) is Scalene triangle
If true then enter $1$ and if false then enter $0$

$Let\quad the\quad points\quad A\left( { x }_{ 1 },{ y }_{ 1 } \right) =\left( 5,6 \right) ,\quad B\left( { x }_{ 2 },{ y }_{ 2 } \right) =\left( 4,2 \right) ,\quad C\left( { x }_{ 3 },{ y }_{ 3 } \right) =\left( 7,5 \right) \\ be\quad the\quad vertices\quad of\quad the\quad \Delta ABC.\\ AB=\sqrt { { \left( { x }_{ 2 }-{ x }_{ 1 } \right) }^{ 2 }+{ \left( { y }_{ 2 }-{ y }_{ 1 } \right) }^{ 2 } } =\sqrt { { \left( 4-5 \right) }^{ 2 }+{ \left( 2-6 \right) }^{ 2 } } units=\sqrt { 17 } units.\\ BC=\sqrt { { \left( { x }_{ 3 }-{ x }_{ 2 } \right) }^{ 2 }+{ \left( { y }_{ 3 }-{ y }_{ 2 } \right) }^{ 2 } } =\sqrt { { \left( 7-4 \right) }^{ 2 }+{ \left( 5-2 \right) }^{ 2 } } units=\sqrt { 18 } units.\\ CA=\sqrt { { \left( { x }_{ 3 }-{ x }_{ 1 } \right) }^{ 2 }+{ \left( { y }_{ 3 }-{ y }_{ 1 } \right) }^{ 2 } } =\sqrt { { \left( 5-7 \right) }^{ 2 }+{ \left( 5-6 \right) }^{ 2 } } units=\sqrt { 5 } units.\\ AB\neq BC\neq CA.\\ So\quad \Delta ABC\quad is\quad a\quad scalene\quad one.\\ Ans-\quad ABC\quad is\quad a\quad scalene\quad \Delta \quad .$

Let the vertices of a triangle ABC be (4, 4), (3, 5), $(-1, -1)$ , then show that $\Delta ABC$ is right angled.

Let

$A\equiv (4, 4), B\equiv (3, 5), C\equiv (-1, -1)$

$AB=\sqrt{(3-4)^{2}+(5-4)^{2}}=\sqrt{2}$

$BC=\sqrt{(-1-3)^{2}+(-1-5)^{2}}=\sqrt{52}$

$AC=\sqrt{(-1-4)^{2}+(-1-4)^{2}}=\sqrt{50}$

$\therefore AB^{2}+AC^{2}=BC^{2}$
Hence by the Pythagoras Theorem, ABC is a right angled triangle.

Show that three points $A (1, -2), B (3, 4) and C (4, 7)$ are lie an a straight line.

$m_{1}=slope of AB=\frac{4-(-2)}{3-1}=3$

$m_{2}=slope of BC=\frac{7-4}{4-3}=3$

$\therefore m_{1}=m_{2}$

$\therefore$ AB is parallel to BC and B is common to both AB and BC. Hence, the points. A

$(1, -2), B(3, 4) and C(4, 7)$ are collinear.
i.e., A, B, C lie on a straight line.

Name the quadrilateral formed, if any by the following points, and give reaons for your answer.
(i) $(-1, -2)$ , $(1, 0)$ , $(-1, 2)$ , $(-3, 0)$
(ii) $(-3, 5)$ , $(3, 1)$ , $(0, 3)$ , $(-1, -4)$
(iii) $(4,5)$ , $(7, 6)$ , $(4, 3)$ , $(1, 2)$

Given,

$A(-1, -2)$ ,

$B(1, 0)$ ,

$C(-1, 2)$ ,

$D(-3, 0)$ Determine distances:

$AB$ ,

$BC$ ,

$CD$ ,

$DA$ ,

$AC$ and

$BD$ .

$AB=\sqrt { (1+1)^{ 2 }+(0+2)^{ 2 } } =\sqrt { 4+4 } =\sqrt { 8 } =2\sqrt { 2 }$

$BC=\sqrt { (-1-1)^{ 2 }+(2-0)^{ 2 } } =\sqrt { 4+4 } =\sqrt { 8 } =2\sqrt { 2 }$

$CD=\sqrt { (-3+1)^{ 2 }+(0-2)^{ 2 } } =\sqrt { 4+4 } =\sqrt { 8 } =2\sqrt { 2 }$

$DA=\sqrt { (-1+3)^{ 2 }+(-2-0)^{ 2 } } =\sqrt { 4+4 } =\sqrt { 8 } =2\sqrt { 2 }$

$AB = BC = CD = DA$
Since, the sides of the quadrilateral are equal.

$AC=\sqrt { (-1+1)^{ 2 }+(2+2)^{ 2 } } =\sqrt { 0+16 } =4$

$BD=\sqrt { (-3-1)^{ 2 }+(0-0)^{ 2 } } =\sqrt { 16+0 } =4$
Diagonals of the quadrilateral are also equal.

$\therefore$
we conclude that

$ABCD$ is a square.

Find the distance of the point $(-1, 1)$ from the line $\displaystyle 12(x+6)= 5(y-2)$

The given equation of the line is

$\displaystyle 12(x+6)= 5(y-2)$

$\displaystyle \Rightarrow 12x+72= 5y-10$

$\displaystyle \Rightarrow 12x-5y+82= 0$
Therefore, the distance of point

$(-1,1)$ from the given line

$\displaystyle = \frac{\left | 12(-1)+(-5)(1)+82 \right |}{\sqrt{(12)^{2}+(-5)^{2}}}$ units

$\displaystyle = \frac{\left | -12-5+82 \right |}{\sqrt{169}}$ units

$\displaystyle = \frac{\left | 65 \right |}{13}$ units

$\displaystyle = 5$ units

Find the value of x such that the distance between the points $(2, 5)$ and $(x, -7)$ is $13$ .

$d=13, (x_1, y_1)=(2, 5)$ and

$(x_2, y_2)=(x, 7)$

$d=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}$

$13=\sqrt{(x-2)^2+(-7-5^2)^2}=\sqrt{(x-2)^2+144}$
Squaring both sides,

$13^2=x^2-4x+148$

$169=x^2-4x+148$

$x^2-4x+148-169=0$

$x^2-4x-21=0$

$(x-7)(x+3)=0$

$\therefore x=7$ or

$x=-3$ .

Find the co-ordinate of points on x-axis which are at a distance of $5$ units form the point $(6, -3)$ .

Any point on X-axis has co-ordinates of form

$(x, 0)$

$\therefore (x_1, y_1)=(6, -3)$ and

$(x_2, y_2)=(x, 0)$ and

$d=5$

$\therefore d=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}$

$5=\sqrt{(x-6)^2+[0-(-3)]^2}=\sqrt{(x-6)^2+3^2}$
Squaring on both sides

$5^2(\sqrt{(x-6)^2+9})^2$

$25=x^2-12x+36+9$

$x^2-12x+45-25=0\Rightarrow x^2-12x+20=0\Rightarrow (x-10)(x-2)=0$

$\therefore x-10=0$ or

$x-2=0\Rightarrow x=10$ or

$x=2$

$\therefore$ Required points on X-axis are

$(10, 0)$ and

$(2, 0)$

Show that the triangle whose vertices are $(8, -4), (9, 5)$ and $(0, 4)$ is an isosceles triangle.

We know that the distance between the two points

$(x_1,y_1)$ and

$(x_2,y_2)$ is

$d=\sqrt { { ({ x }_{ 2 }-{ x }_{ 1 }) }^{ 2 }+{ ({ y }_{ 2 }-{ y }_{ 1 }) }^{ 2 } }$

Let the given vertices be

$A=(8,-4)$ ,

$B=(9,5)$ and

$C=(0,4)$

We first find the distance between

$A=(8,-4)$ and

$B=(9,5)$ as follows:

$AB=\sqrt { { ({ x }_{ 2 }-{ x }_{ 1 }) }^{ 2 }+{ ({ y }_{ 2 }-{ y }_{ 1 }) }^{ 2 } } =\sqrt { { (10-2) }^{ 2 }+{ (1-1) }^{ 2 } } =\sqrt { { 8 }^{ 2 }+{ 0 }^{ 2 } } =\sqrt { 64 } =8$

Similarly, the distance between

$B=(9,5)$ and

$C=(0,4)$ is:

$BC=\sqrt { { ({ x }_{ 2 }-{ x }_{ 1 }) }^{ 2 }+{ ({ y }_{ 2 }-{ y }_{ 1 }) }^{ 2 } } =\sqrt { { (6-10) }^{ 2 }+{ (9-1) }^{ 2 } } =\sqrt { { (-4) }^{ 2 }+{ 8 }^{ 2 } } =\sqrt { 16+64 } =\sqrt { 80 } =\sqrt { 4^{ 2 }\times 5 } =4\sqrt { 5 }$

Now, the distance between

$C=(0,4)$ and

$A=(8,-4)$ is:

$CA=\sqrt { { ({ x }_{ 2 }-{ x }_{ 1 }) }^{ 2 }+{ ({ y }_{ 2 }-{ y }_{ 1 }) }^{ 2 } } =\sqrt { { (6-2) }^{ 2 }+{ (9-1) }^{ 2 } } =\sqrt { { 4 }^{ 2 }+{ 8 }^{ 2 } } =\sqrt { 16+64 } =\sqrt { 80 } =\sqrt { 4^{ 2 }\times 5 } =4\sqrt { 5 }$

We also know that If any two sides have equal side lengths, then the triangle is isosceles.

Here, since the lengths of the two sides are equal that is

$BC=CA=4\sqrt {5}$

Hence, the given vertices are the vertices of an isosceles triangle.

Prove that the points $A(-3, -2), B(5, -2), C(9, 3)$ and $D(1, 3)$ are the vertices of a parallelogram.

We know that the distance between the two points

$(x_1,y_1)$ and

$(x_2,y_2)$ is

$d=\sqrt { { ({ x }_{ 2 }-{ x }_{ 1 }) }^{ 2 }+{ ({ y }_{ 2 }-{ y }_{ 1 }) }^{ 2 } }$

Let the given vertices be

$A=(-3,-2)$ ,

$B=(5,-2)$ ,

$C=(9,3)$ and

$D=(1,3)$

We first find the distance between

$A=(-3,-2)$ and

$B=(5,-2)$ as follows:

$AB=\sqrt { { ({ x }_{ 2 }-{ x }_{ 1 }) }^{ 2 }+{ ({ y }_{ 2 }-{ y }_{ 1 }) }^{ 2 } } =\sqrt { { (1-(-5)) }^{ 2 }+{ (-11-(-3)) }^{ 2 } } =\sqrt { { (1+5) }^{ 2 }+{ (-11+3) }^{ 2 } } =\sqrt { { 6 }^{ 2 }+{ (-8) }^{ 2 } }$

$=\sqrt { 36+64 } =\sqrt { 100 } =\sqrt { 10^{ 2 } } =10$

Similarly, the distance between

$B=(5,-2)$ and

$C=(9,3)$ is:

$BC=\sqrt { { ({ x }_{ 2 }-{ x }_{ 1 }) }^{ 2 }+{ ({ y }_{ 2 }-{ y }_{ 1 }) }^{ 2 } } =\sqrt { { (7-1) }^{ 2 }+{ (-6-(-11)) }^{ 2 } } =\sqrt { { 6 }^{ 2 }+{ (-6+11) }^{ 2 } } =\sqrt { { 6 }^{ 2 }+{ 5 }^{ 2 } } =\sqrt { 36+25 }$

$=\sqrt { 61 }$

Now, the distance between

$C=(9,3)$ and

$D=(1,3)$ is:

$CD=\sqrt { { ({ x }_{ 2 }-{ x }_{ 1 }) }^{ 2 }+{ ({ y }_{ 2 }-{ y }_{ 1 }) }^{ 2 } } =\sqrt { { (1-7) }^{ 2 }+{ (2-(-6)) }^{ 2 } } =\sqrt { { (-6) }^{ 2 }+{ (2+6) }^{ 2 } } =\sqrt { { (-6) }^{ 2 }+{ 8 }^{ 2 } } =\sqrt { 36+64 }$

$=\sqrt { 100 } =\sqrt { 10^{ 2 } } =10$

Now, the distance between

$D=(1,3)$ and

$A=(-3,-2)$ is:

$DA=\sqrt { { ({ x }_{ 2 }-{ x }_{ 1 }) }^{ 2 }+{ ({ y }_{ 2 }-{ y }_{ 1 }) }^{ 2 } } =\sqrt { { (1-(-5)) }^{ 2 }+{ (2-(-3)) }^{ 2 } } =\sqrt { { (1+5) }^{ 2 }+{ (2+3) }^{ 2 } } =\sqrt { { 6 }^{ 2 }+{ 5 }^{ 2 } } =\sqrt { 36+25 }$

$=\sqrt { 61 }$

We also know that if the opposite sides have equal side lengths, then

$ABCD$ is a parallelogram.

Here, since the lengths of the opposite sides are equal that is:

$AB=CD=10$ and

$BC=DA=\sqrt {61}$

Hence, the given vertices are the vertices of parallelogram.

Find the distance between the points $P(3, 4, -7)$ and $Q(-2, 5, 10)$

Distance between two points

$\sqrt { { (x2-x1) }^{ 2 }+{ (y2-y1) }^{ 2 }+{ (z2-z1) }^{ 2 } }$

$=\sqrt { { (5) }^{ 2 }+{ (-1) }^{ 2 }+{ (-17) }^{ 2 } }$

$=\sqrt {{25}+{1}+{289}}$

$=\sqrt {315} \quad Ans$

The center of a circle is $(2a-1,7)$ and it passes through the point $(-3,-1)$ . If the diameter of the circle is $20$ units, then find the value of $a$ .

Centre:

$C(2a-1,7)$

Point:

$P(-3,-1)$

Radius=

$\cfrac { 1 }{ 2 } (20)=10$

As circle passes through P, distance between p and centre is equal to radius.

${ (PC) }^{ 2 }={ (10 })^{ 2 }\\ =>(2a-1+3)^{ 2 }+\left( 7+1 \right) ^{ 2 }=(10)^{ 2 }\\ =>({ a+1 })^{ 2 }=\cfrac { 36 }{ 4 } \\ =>({ a+1 })^{ 2 }=9\\ =>a=2,-4$

The point $(1, 2.5)$ divides the line in two equal parts, joining the points $(a, 1)$ and $(11, 4)$ . Find the value of $\mid a \mid$ .

Mid point formula

$\left ( \dfrac{x_{1}+x_{2}}{2},\dfrac{y_{1}+y_{2}}{2} \right )$

$1=\dfrac{a+11}{2}$

$a=-9$

$\mid a \mid = 9$

Find the circumcentre and circumradius of the triangle whose vertices are $(1,1), (2,-1)$ and $(3,2)$

Let the coordinates of the vertices of the triangle are

$A(1,1), B(2,-1)$ and

$C(3,2)$

Now, Lets find the distance between each of the three points.

We will use the distance formula to find the distance between

$AB, BC$ and

$AC$

Thus

$AB= \sqrt{(2-1)^2+ (-1-1)^2}=\sqrt{5}$

$BC= \sqrt{(3-2)^2+ (2+1)^2}=\sqrt{10}$

$AC=\sqrt{(3-1)^2+ (2-1)^2}=\sqrt{5}$

Now

$AB^2+AC^2= 10 =BC^2$

Thus it form a right angled triangle, right angled at

$A.$

Now the circumcentre will be the mid-point of the hypotenuse

$BC.$

Thus mid-point of

$BC$ is

$\left (\dfrac{2+3}{2},\dfrac{-1+2}{2} \right ) = \left (\dfrac{5}{2},\dfrac{1}{2} \right ).$

And the circumradius is the half the length of

$BC$

$\Rightarrow$ Circumradius

$= \dfrac{\sqrt{10}}{2}$

The vertices of a triangle are $A(1,1), B(4,5)$ and $C(6,13)$ . Find $\cos{A}$

Using law of cosines, we have,

$a^2=b^2+c^2-2bccosA$

From the given points,

$a^2=BC^2=68$ ,

$b^2=AC^2=169$ ,

$c^2=AB^2=25$

Substituting these values, we have

$cosA=\dfrac{b^2+c^2-a^2}{2bc}$

$cosA=\dfrac{169+25-68}{2 \times 13 \times 5}$ =

$\dfrac{126}{130}$

$=\dfrac{63}{65}$

If points $(3,K)$ and $(K,5)$ are equidistant from a point $(0,2)$ , then find the value of $K$ .

$d=[(3-0)^2+(k-2)^2]^{1/2}$ ……..

$(1)$

$d=[(k-0)^2+(5-2)^2]^{1/2}$ ………..

$(2)$

$(1) = (2)$

Squaring both sides

$3^2+(k-2)^2=k^2+3^2$

Take square root both sides

$k-2=\pm k$

$(+)$

$k-2=k$

$k-2=0$ not possible

$(-)$

$k-2=-k$

$k=1$ .

1381387_1211209_ans_e7aa1a8fb4604be8866d6e50dcd3a7fb.PNG

Find the angles between the lines
$x-\sqrt{3}y=1$

Consider the given equations,

$x-\sqrt{3}y=1$ …..(1)

$\sqrt{3}x-y=1$ …..(2)

From equation 1^st,

$y=\dfrac{1}{\sqrt{3}}x-\dfrac{1}{\sqrt{3}}$

Compare with, $y={{m}_{1}}x+C$

${{m}_{1}}=\dfrac{1}{\sqrt{3}}$

From equation 2^nd ,

$y=\sqrt{3}x-1$

Compare with $y={{m}_{2}}x+C$

${{m}_{2}}=\sqrt{3}$

Now,let the angle between line 1^st and 2^{nd $=\theta$},

Using the formula, $\tan \theta =\dfrac{{{m}_{2}}-{{m}_{1}}}{1+{{m}_{1}}{{m}_{2}}}$

$\tan \theta =\dfrac{\sqrt{3}-\dfrac{1}{\sqrt{3}}}{1+\sqrt{3}\times \dfrac{1}{\sqrt{3}}}=\dfrac{\sqrt{3}\times \sqrt{3}-1}{1+1}$

$\tan \theta =\dfrac{3-1}{2}=1$

$\tan \theta =1$

$\tan \theta =\tan \dfrac{\pi }{4}$

$\theta =\dfrac{\pi }{4}$

Hence, this is the answer.

The corner point of the feasible region determined by the system of Iinear constraints are $( 0,10 )$ , $( 5,5 ) , ( 15,15 )$ and $( 0,20 )$ .Let $Z = p x + q y ,$ where $p , q > 0$ .Find the condition on $p$ and $q$ so that the maximum of $Z$ occurs at both the points $( 15,15 )$ and $( 0,20 )$ .

Given

$Z=p{x}+q{y}$

$Z$ is maximum at both points

$\implies Z$ at

$(15,15)$ is equal to

$Z$ at

$(0,20)$

$\implies 15{p}+15{q}=20{q}\implies 3{p}=q$

If the points $A(4, 3)$ and $B(x, 5)$ are on the circle with the centre $O(2, 3)$ . Find the value of $x$ .

Let equation of circle be

$x^2+y^2+2gx+2fy+c=0$

$(-g,-f)\rightarrow$ centre is

$(2,3)$

$x^2+y^2+2(-2)(x)+2(-3)y+c=0$

$A(4,3)$ lies on circle.

$(4)^2+(3)^3-4(4)-6(3)+c=0$

$c=9$

$\therefore x^2+y^2-4x-6y+9=0$ is equation of circle.

$B(x,5)$

$x^2+25-4x-6(5)+9=0$

$x^2-4x+4=0$

$(x-2)^2=0\Rightarrow x=2$

Find the point on the $x-$ axis which is equidistant from the points $\left(-3,4\right)$ and $\left(2,5\right)$

1338031_1276082_ans_ab36789e760c4aa8a422642b15d69b5f.PNG

$AD$ is the median of $\triangle ABC$ and bisectors of $\angle ADB$ and $\angle ADC$ are $DE$ and $DF$ , which meet $AB$ at $E$ and $AC$ at $F$ . Prove that $EF\parallel BC$ .

$\Delta DAE, \, DE$ bisect angle

$ADB$

So we have

$\dfrac{DA}{DB} = \dfrac{AE }{EB}$ ... (1)

Similarly in

$\Delta DAC, \, DE$ bisect angle

$ADC$

we get

$\dfrac{DA}{DC} = \dfrac{AF}{AC}$

$(DC = DB)$

$\dfrac{DA}{DB} = \dfrac{AF}{FC}$ ... (2)

from eq (1) and eq (2)

$\dfrac{AE}{EB} = \dfrac{AF}{FC}$

$\Delta ABC$

$EF || BE (... BPT)$

The equation of the straight line passing through the point of intersection of the straight lines $\dfrac{x}{a} =\dfrac{y}{b} = 1$ and $\dfrac{x}{b} + \dfrac{y}{a} = 1$ and having infinite slopes is

Let

$L_{1}:\dfrac{x}{a}+\dfrac{y}{b}-1=0$ &

$L_{2}=\dfrac{x}{b}+\dfrac{y}{a}-1=0$

the required line would be of the form

$L_{1}-\lambda L_{2}=0$

$\Rightarrow (\dfrac{x}{a}+\dfrac{y}{b}-1)-\lambda (\dfrac{x}{b}+\dfrac{y}{a}-1)=0$

$\Rightarrow x(\dfrac{1}{a}-\dfrac{\lambda }{b})+y(\dfrac{1}{b}-\dfrac{\lambda }{a})-1+\lambda =0$

As slope =

$\infty \Rightarrow$ line is parallel to y-axis

$\Rightarrow$ coeff. of

$y=0\Rightarrow \dfrac{\lambda }{a}=\dfrac{1}{b}\Rightarrow \lambda =\dfrac ab$

$\Rightarrow Eq^{n}$ of line :

$x(\dfrac{1}{a}-\dfrac{a}{b^{2}})+y(\dfrac{1}{b}-\dfrac{1}{b})-1+\dfrac{a}{b}=0$

$\Rightarrow x(\dfrac{b^{2}-a^{2}}{ab^{2}})+\dfrac{a}{b}-1=0$

Given the points $A\left(0,\,4\right)$ and $B\left(0,\,-4\right)$ , find the locus of $P\left(x,y\right)$ such that $\left|AP-BP\right|=6$

1338047_1276691_ans_355d56cf004547b0ada9a70ced638e17.PNG

If $\angle B$ and $\angle Q$ are acute angels such that $\sin B-\sin Q$ , then prove that $\angle B=\angle Q$ .

We have,

In $\Delta ABC\,\,and\,\,\Delta PQR$

Given that, $\angle B\,\,and\,\,\angle Q$ are acute angles

$\therefore AC=k\times PR\,\,and\,\,AB=k\times PQ$

From $\Delta ACB\,$

By Pythagoras theorem,

$A{{B}^{2}}=A{{C}^{2}}+B{{C}^{2}}$

$\Rightarrow {{\left( k\times PR \right)}^{2}}={{\left( k\times PQ \right)}^{2}}+B{{C}^{2}}$

$\Rightarrow {{k}^{2}}\times P{{R}^{2}}={{k}^{2}}\times P{{Q}^{2}}+B{{C}^{2}}$

$\Rightarrow B{{C}^{2}}={{k}^{2}}\times P{{R}^{2}}-{{k}^{2}}\times P{{Q}^{2}}$

$\Rightarrow B{{C}^{2}}={{k}^{2}}\left[ P{{R}^{2}}-P{{Q}^{2}} \right]$

$\Rightarrow BC=k\sqrt{P{{R}^{2}}-P{{Q}^{2}}}$

From right angle triangle PRQ,

Using Pythagoras theorem,

We have,

$P{{Q}^{2}}=P{{R}^{2}}+Q{{R}^{2}}$

$\Rightarrow Q{{R}^{2}}=P{{Q}^{2}}-P{{R}^{2}}$

Hence,

$\Delta ACB\sim \Delta PRQ\,\,\left( S.S.S.\,\,similarity\,triangle \right)$

$\therefore \,\,\angle B=\angle Q$

Hence, this is the answer.

Reduce the equation in to slope intercept from $6x+3y-5=$

$6x+3y-5=0$

$3y=-6x+5$

$y=\dfrac{-6}{3}x+\dfrac{5}{3}$

$\Rightarrow y=-2x+\dfrac{5}{3}$

This is the slope intercept form

$(y=mx+c)$

where slope

$=-2$ & intercept

$=\dfrac{5}{3}$ .

Show that the points
A(2,-2), B(8,4) , C(5,7) ,D(-1,1) are the vertices of a rectangle .

We know the distance between the two points is given by,

$d=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}$

Here

$AB=\sqrt{(8-2)^2+(4+2)^2}$

$=\sqrt{36+36}$

$=6\sqrt{2},$

$BC=\sqrt{(5-8)^2+(7-4)^2}=\sqrt{9+9}$

$=3\sqrt{2},$

$CD=\sqrt{(-1-5)^2+(1-7)^2}=\sqrt{36+36}$

$=6\sqrt{2}$ and

$AD=\sqrt{(2+1)^2+(-2-1)^2}=\sqrt{9+9}$

$=3\sqrt{2}$

$AC=\sqrt{(5-2)^2+(7+2)^2}=\sqrt{90}$ and

$BD=\sqrt{(-1-8)^2+(1-4)^2}=\sqrt{90}$

Here

$AB=CD,BC=AD,AC=BD$

i.e.,opposite sides are equal and diagonal are equal.

Hence it is a rectangle

Find the value of $y$ for which the distance between the points $P(2, -3)$ and $Q(10, y)$ is $10$ units.

Given points are

$P(2,-3)$ and

$Q(10,y)$

Distance between two point

$(x_1,y_1)$ and

$(x_2,y_2)$ is given by,

$=\sqrt{ { \left( {x_2}-{ { x_1 } } \right) }^{ 2}+{ \left( {y_2 }-{ y_1 } \right) }^{ 2 } }$

$PQ = \sqrt{8^{2}+(y+3)^{2}} =10$

$\Rightarrow \sqrt{64+y^{2}+9+6y} = 10$

$\Rightarrow y^{2}+6y+73=100$

$\Rightarrow y^{2}+6y-27=0$

$\Rightarrow y^{2}+9y-3y-27=0$

$\Rightarrow y(y+9)-3(y+9)=0$

$\Rightarrow (y+9)(y-3)=0$

$\Rightarrow y=-9,3$

Prove that the points $A(1,-3),B(-3,0)$ and $C(4,1)$ are the vertices of a right angled Isosceles triangle

$AB=\sqrt{(1-(-3))^2+(-3-0)^2}=\sqrt{4^2+3^2}=\sqrt{16+9}=\sqrt{25}=5$

$BC=\sqrt{(-3-4)^2+(0-1)^2}=\sqrt{49+1}=\sqrt{50}=5\sqrt{2}$

$AC=\sqrt{(1-4)^2+(-3-1)^2}=\sqrt{9+16}=\sqrt{25}=5$

$AB^2+AC^2=5^2+5^2=50=BC^2$

By Pythagoras theorem

$\angle A=90^0$

and

$AC=AB$

$\therefore ABC$ is right isosceles

$\Delta$

2040490_1481731_ans_7755eb7e1d1f4aa8832a500b5bdad53d.png

Using the method of slope, show that the following points are collinear:
$A(16, -18), B(3,-6), C(-10,6)$

1414166_1667226_ans_7eef954a3b8a4a3389e1ab4e728f7142.jpg

Find the slope of a line whose inclination is
(i) $45^o$
(ii) $30^o$

The slope of a line inclination :

(i)

$45^o$

$Slope = tan \,45^o = 1$

(ii)

$30^o$

$Slope = tan\,30^o = 1/\sqrt{3}$

Prove that the line through $\left(0,0\right)$ and $\left(2,3\right)$ is parallel to the line through $\left(2, -2 \right)$ and $\left(6,4\right)$ .

Let the slope of the line through

$\left(2,-2\right)$ and

$\left(6,4\right)$ be

$m_1$

So,

$m_1 = \dfrac{\left(y_2 - y_1\right)}{\left (x_2 - x_1\right)}$

$=\dfrac{\left(4 + 2\right)}{\left (6 - 2\right)}$

$=\dfrac{6}{4} = \dfrac{3}{2}$

Let the slope of the line through

$\left(0,0\right)$ and

$\left(2,3\right)$ be

$m_2$

So,

$m_2 = \dfrac{\left(y_2 - y_1\right)}{\left (x_2 - x_1\right)}$

$=\dfrac{\left(3 -0\right)}{\left (2 - 0\right)}$

$= \dfrac{3}{2}$

It's clearly seen that the slopes

$m_1 = m_2$

Thus, the lines are parallel to each other.

Show that the triangle formed by the point $A \left(1,3\right) \, B \left(3, -1\right)\, and \, C \left(-5, -5 \right)$ is a right - angled triangle by using slopes.

Given, points

$A \left(1,3\right) , B \left(3,-1\right) \, and \, C \left(-5, -5\right)$ form a triangle

Now,

Slope of the line

$AB = m_1 = \dfrac{\left(-1 - 3\right)}{\left (3 - 1\right)} = \dfrac{-4}{2} = -2$

And,

Slope of the line

$BC = m_2 = \dfrac{\left(-5 + 1\right)}{\left (-5 - 3\right)} = \dfrac{-4}{-8} = \dfrac{1}{2}$

Hence,

$m_1 \times m_2 = \left(-2\right) \times \left(\dfrac{1}{2}\right) = -1$

So, the line AB and BC are perpendicular to each other .

Therefore,

$\triangle ABC$ is a right - angled triangle.

Prove that the line through $\left(-2,6\right)$ and $\left(4, 8 \right)$ is perpendicular to the line through $\left(8,12\right)$ and $\left(4,24\right)$

Let the slope of the line through points

$\left (-2,6\right)$ and

$\left(4,8\right)$ be

$m_1$

So,

$m_1 = \dfrac{\left(y_2 - y_1\right)}{\left (x_2 - x_1\right)}$

$= \dfrac{\left(8 - 6\right)}{\left (4 + 2\right)}$

$= \dfrac{2}{6}$

$= \dfrac{1}{3}$

And, let the slope of the line through

$\left(8,2\right)$ and

$\left(4,24\right)$ be

$m_2$

So,

$m_2 = \dfrac{\left(y_2 - y_1\right)}{\left (x_2 - x_1\right)}$

$= \dfrac{\left(24 - 12\right)}{\left (4 - 8\right)}$

$= \dfrac{12}{\left(-4\right)}$

$= -3$

Now, product of slopes is

$m_1 \times m_2 = \dfrac{1}{3} \times \left(-3\right) = -1$

Thus, the lines are perpendicular to each other.

Prove that the line through $\left(0,0\right)$ and $\left(2,3\right)$ is parallel to the line through $\left(2, -2 \right)$ and $\left(6,4\right)$ .

Let the slope of the line through

$\left(0,0\right)$ and

$\left(2,3\right)$ be

$m_1$
So,

$m_1 = \dfrac{\left(y_2 - y_1\right)}{\left (x_2 - x_1\right)}$

$=\dfrac{\left(3-0\right)}{\left (2-0\right)}=\dfrac{3}{2}$
Similarly

$m_{2}$

$=\dfrac{\left(4 + 2\right)}{\left (6 - 2\right)}$

$=\dfrac{6}{4} = \dfrac{3}{2}$
It's clearly seen that the slopes

$m_1 = m_2$
Thus, the lines are parallel to each other.

The lien through the point (h,3) and (4,1) intersects the line $7 x = 9y = 19$ at right angle. Find the value of h

Given line is

$7 x - 9y = 19$ which is perpendicular to the our line

given point are (h,3),(4,1)

For perpendicular lines

$m_1m_2=-1$ where

$m_1,m_2$ are slope of perpendicular lines

$\therefore m_1=\frac{-1}{\frac{-7}{-9}}\Rightarrow m_1=\frac{9}{7}$

Therefore,Required equations is of the form

$y - y_{1} = m_1 ( x - x_{1})$

$y-1=\frac{9}{7}(x-4)$

$7y-7=9x-36$

$9x-7y-29=0$

Required equation,For h

$\Rightarrow 9\times h -7\times 3-29=0$

$h=\frac{50}{3}$

Answer

Find slope of a line passing through the points $A(3, 1)$ and $B(5, 3).$

1895683_1911511_ans_84fe5d780bc5412cbe41b542a292afd6.jpeg

Show that the line given by $x(a+2b)+y(a+3b)=a+b$ for different values of $a$ and $b$ pass through a fixed point.

$\begin{array}{l} x\left( { a+zb } \right) +y\left( { a+zb } \right) =a+b \\ a\left( { x+y } \right) +b\left( { 2x+3y } \right) =a+b \\ x+y=1\to (i) \\ 2x+3y=1\to (ii) \\ Multiply\, \, 2\, \, in\, \, eq.\, \, (i) \\ 2x+2y=2 \\ 2x+3y=1 \\ -\, \, \, \, \, -\, \, \, \, \, \, \, - \\ \overline { \, \, \, \, \, \, \, \, -y=1 } \\ y=-1 \\ put\, \, y\, \, in\, \, eq\, (i) \\ x-1=1 \\ x=2 \\ fixed\, \, po{ { int } }\, \, \left( { 2,-1 } \right) \end{array}$

Straight Lines - Class 11 Commerce Maths - Extra Questions

Find the slope of the line passing through the points A(2,3)A(2, 3) and B(4,7)B(4, 7).

Find the slope of line having inclination 60o{60}^{o}.

Find the slope of the line passing through the points C(3,5)C(3, 5) and D(−2,−3)D(-2, -3)

If the straight line joining two points P(5,8)P(5,8) and Q(8,k)Q (8,k) is parallel to x−axisx-axis, then write the value of k k.

If the distance between the points (4,p)(4,\,p) and (1,0)(1,\,0) is 55, then the find the value of pp.

The points (−2,5)(-2,5) and (3,−5)(3,-5) are plotted in xy planes. Find the slope and y intercept of the line joining the points.

Find the distance between (−4,5)\left(-4,5\right) from the origin.

Solve the following question:Find the slope of the line passing through the points A(2,3)A(2, 3) and B(4,7)B(4, 7).

The type of triangle formed by the points A (5, 6), B (4, 2) and C (7, 5) is Scalene triangleIf true then enter 11 and if false then enter 00

Let the vertices of a triangle ABC be (4, 4), (3, 5), (−1,−1) (-1, -1), then show that ΔABC\Delta ABC is right angled.

Show that three points A(1,−2),B(3,4)andC(4,7)A (1, -2), B (3, 4) and C (4, 7) are lie an a straight line.

Name the quadrilateral formed, if any by the following points, and give reaons for your answer.(i) (−1,−2)(-1, -2), (1,0)(1, 0), (−1,2)(-1, 2), (−3,0)(-3, 0)(ii)(−3,5)(-3, 5), (3,1)(3, 1), (0,3)(0, 3), (−1,−4)(-1, -4)(iii)(4,5)(4,5), (7,6)(7, 6), (4,3)(4, 3), (1,2)(1, 2)

Find the distance of the point (−1,1)(-1, 1) from the line 12(x+6)=5(y−2)\displaystyle 12(x+6)= 5(y-2)

Find the value of x such that the distance between the points (2,5)(2, 5) and (x,−7)(x, -7) is 1313.

Find the co-ordinate of points on x-axis which are at a distance of 55 units form the point (6,−3)(6, -3).

Show that the triangle whose vertices are (8,−4),(9,5)(8, -4), (9, 5) and (0,4)(0, 4) is an isosceles triangle.

Prove that the points A(−3,−2),B(5,−2),C(9,3)A(-3, -2), B(5, -2), C(9, 3) and D(1,3)D(1, 3) are the vertices of a parallelogram.

Find the distance between the points P(3,4,−7)P(3, 4, -7) and Q(−2,5,10)Q(-2, 5, 10)

The center of a circle is (2a−1,7)(2a-1,7) and it passes through the point (−3,−1)(-3,-1). If the diameter of the circle is 2020 units, then find the value of aa.

The point (1,2.5)(1, 2.5) divides the line in two equal parts, joining the points (a,1)(a, 1) and (11,4)(11, 4). Find the value of ∣a∣\mid a \mid.

Find the circumcentre and circumradius of the triangle whose vertices are (1,1),(2,−1)(1,1), (2,-1) and (3,2)(3,2)

The vertices of a triangle are A(1,1),B(4,5)A(1,1), B(4,5) and C(6,13)C(6,13). Find cosA\cos{A}

If points (3,K)(3,K) and (K,5)(K,5) are equidistant from a point (0,2)(0,2), then find the value of KK.

Find the angles between the linesx−√3y=1x-\sqrt{3}y=1

If the points A(4,3)A(4, 3) and B(x,5)B(x, 5) are on the circle with the centre O(2,3)O(2, 3). Find the value of xx.

Find the point on the x−x-axis which is equidistant from the points (−3,4)\left(-3,4\right) and (2,5)\left(2,5\right)

ADAD is the median of △ABC\triangle ABC and bisectors of ∠ADB\angle ADB and∠ADC\angle ADC are DEDE and DFDF, which meet ABAB at EE and ACAC at FF. Prove that EF∥BCEF\parallel BC.

The equation of the straight line passing through the point of intersection of the straight linesxa=yb=1\dfrac{x}{a} =\dfrac{y}{b} = 1 and xb+ya=1\dfrac{x}{b} + \dfrac{y}{a} = 1 and having infinite slopes is

Given the points A(0,4)A\left(0,\,4\right) and B(0,−4)B\left(0,\,-4\right), find the locus of P(x,y)P\left(x,y\right) such that |AP−BP|=6\left|AP-BP\right|=6

If ∠B\angle B and ∠Q\angle Q are acute angels such that sinB−sinQ\sin B-\sin Q, then prove that ∠B=∠Q\angle B=\angle Q.

Reduce the equation in to slope intercept from 6x+3y−5=6x+3y-5=

Show that the points A(2,-2), B(8,4) , C(5,7) ,D(-1,1) are the vertices of a rectangle .

Find the value of yy for which the distance between the points P(2,−3)P(2, -3) and Q(10,y)Q(10, y) is 1010 units.

Prove that the points A(1,−3),B(−3,0)A(1,-3),B(-3,0) and C(4,1)C(4,1) are the vertices of a right angled Isosceles triangle

Using the method of slope, show that the following points are collinear:A(16,−18),B(3,−6),C(−10,6)A(16, -18), B(3,-6), C(-10,6)

Find the slope of a line whose inclination is (i) 45o45^o(ii) 30o30^o

Prove that the line through (0,0)\left(0,0\right) and (2,3)\left(2,3\right) is parallel to the line through (2,−2)\left(2, -2 \right) and (6,4)\left(6,4\right).

Show that the triangle formed by the point A(1,3)B(3,−1)andC(−5,−5)A \left(1,3\right) \, B \left(3, -1\right)\, and \, C \left(-5, -5 \right) is a right - angled triangle by using slopes.

Prove that the line through (−2,6)\left(-2,6\right) and (4,8)\left(4, 8 \right) is perpendicular to the line through (8,12)\left(8,12\right) and (4,24)\left(4,24\right)

Prove that the line through (0,0)\left(0,0\right) and (2,3)\left(2,3\right) is parallel to the line through (2,−2)\left(2, -2 \right) and (6,4)\left(6,4\right).

The lien through the point (h,3) and (4,1) intersects the line 7x=9y=197 x = 9y = 19 at right angle. Find the value of h

Find slope of a line passing through the points A(3,1)A(3, 1) and B(5,3).B(5, 3).

Show that the line given by x(a+2b)+y(a+3b)=a+bx(a+2b)+y(a+3b)=a+b for different values of aa and bb pass through a fixed point.

Class 11 Commerce Maths Extra Questions

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Find the slope of the line passing through the points $A(2, 3)$ and $B(4, 7)$ .

Find the slope of line having inclination ${60}^{o}$ .

Find the slope of the line passing through the points $C(3, 5)$ and $D(-2, -3)$

If the straight line joining two points $P(5,8)$ and $Q (8,k)$ is parallel to $x-axis$ , then write the value of $k$ .

If the distance between the points $(4,\,p)$ and $(1,\,0)$ is $5$ , then the find the value of $p$ .

The points $(-2,5)$ and $(3,-5)$ are plotted in xy planes. Find the slope and y intercept of the line joining the points.

Find the distance between $\left(-4,5\right)$ from the origin.

Solve the following question:
Find the slope of the line passing through the points $A(2, 3)$ and $B(4, 7)$ .

The type of triangle formed by the points A (5, 6), B (4, 2) and C (7, 5) is Scalene triangle
If true then enter $1$ and if false then enter $0$

Let the vertices of a triangle ABC be (4, 4), (3, 5), $(-1, -1)$ , then show that $\Delta ABC$ is right angled.

Show that three points $A (1, -2), B (3, 4) and C (4, 7)$ are lie an a straight line.

Name the quadrilateral formed, if any by the following points, and give reaons for your answer.
(i) $(-1, -2)$ , $(1, 0)$ , $(-1, 2)$ , $(-3, 0)$
(ii) $(-3, 5)$ , $(3, 1)$ , $(0, 3)$ , $(-1, -4)$
(iii) $(4,5)$ , $(7, 6)$ , $(4, 3)$ , $(1, 2)$

Find the distance of the point $(-1, 1)$ from the line $\displaystyle 12(x+6)= 5(y-2)$

Find the value of x such that the distance between the points $(2, 5)$ and $(x, -7)$ is $13$ .

Find the co-ordinate of points on x-axis which are at a distance of $5$ units form the point $(6, -3)$ .

Show that the triangle whose vertices are $(8, -4), (9, 5)$ and $(0, 4)$ is an isosceles triangle.

Prove that the points $A(-3, -2), B(5, -2), C(9, 3)$ and $D(1, 3)$ are the vertices of a parallelogram.

Find the distance between the points $P(3, 4, -7)$ and $Q(-2, 5, 10)$

The center of a circle is $(2a-1,7)$ and it passes through the point $(-3,-1)$ . If the diameter of the circle is $20$ units, then find the value of $a$ .

The point $(1, 2.5)$ divides the line in two equal parts, joining the points $(a, 1)$ and $(11, 4)$ . Find the value of $\mid a \mid$ .

Find the circumcentre and circumradius of the triangle whose vertices are $(1,1), (2,-1)$ and $(3,2)$

The vertices of a triangle are $A(1,1), B(4,5)$ and $C(6,13)$ . Find $\cos{A}$

If points $(3,K)$ and $(K,5)$ are equidistant from a point $(0,2)$ , then find the value of $K$ .

Find the angles between the lines
$x-\sqrt{3}y=1$

If the points $A(4, 3)$ and $B(x, 5)$ are on the circle with the centre $O(2, 3)$ . Find the value of $x$ .

Find the point on the $x-$ axis which is equidistant from the points $\left(-3,4\right)$ and $\left(2,5\right)$

$AD$ is the median of $\triangle ABC$ and bisectors of $\angle ADB$ and $\angle ADC$ are $DE$ and $DF$ , which meet $AB$ at $E$ and $AC$ at $F$ . Prove that $EF\parallel BC$ .

The equation of the straight line passing through the point of intersection of the straight lines $\dfrac{x}{a} =\dfrac{y}{b} = 1$ and $\dfrac{x}{b} + \dfrac{y}{a} = 1$ and having infinite slopes is

Given the points $A\left(0,\,4\right)$ and $B\left(0,\,-4\right)$ , find the locus of $P\left(x,y\right)$ such that $\left|AP-BP\right|=6$

If $\angle B$ and $\angle Q$ are acute angels such that $\sin B-\sin Q$ , then prove that $\angle B=\angle Q$ .

Reduce the equation in to slope intercept from $6x+3y-5=$

Show that the points
A(2,-2), B(8,4) , C(5,7) ,D(-1,1) are the vertices of a rectangle .

Find the value of $y$ for which the distance between the points $P(2, -3)$ and $Q(10, y)$ is $10$ units.

Prove that the points $A(1,-3),B(-3,0)$ and $C(4,1)$ are the vertices of a right angled Isosceles triangle

Using the method of slope, show that the following points are collinear:
$A(16, -18), B(3,-6), C(-10,6)$

Find the slope of a line whose inclination is
(i) $45^o$
(ii) $30^o$

Prove that the line through $\left(0,0\right)$ and $\left(2,3\right)$ is parallel to the line through $\left(2, -2 \right)$ and $\left(6,4\right)$ .

Show that the triangle formed by the point $A \left(1,3\right) \, B \left(3, -1\right)\, and \, C \left(-5, -5 \right)$ is a right - angled triangle by using slopes.

Prove that the line through $\left(-2,6\right)$ and $\left(4, 8 \right)$ is perpendicular to the line through $\left(8,12\right)$ and $\left(4,24\right)$

Prove that the line through $\left(0,0\right)$ and $\left(2,3\right)$ is parallel to the line through $\left(2, -2 \right)$ and $\left(6,4\right)$ .

The lien through the point (h,3) and (4,1) intersects the line $7 x = 9y = 19$ at right angle. Find the value of h

Find slope of a line passing through the points $A(3, 1)$ and $B(5, 3).$

Show that the line given by $x(a+2b)+y(a+3b)=a+b$ for different values of $a$ and $b$ pass through a fixed point.