Explanation
Distance between two points =√(x2−x1)2+(y2−y1)2Distance between the points A(3,1) and B(−3,2)=√(−3−3)2+(2−1)2=√36+1=√37
Distance between the points B(−3,2) and C(0,2−√3)=√(0+3)2+(2−√3−2)2=√9+3=√12
Distance between the points A(3,1) and C(0,2−√3)
=√(0−3)2+(2−√3−1)2=√9+1+3−2√3=√13−2√3
Since the length of the sides between all vertices are different, they are the vertices of a scalene triangle.
Area of a triangle =|x1(y2−y3)+x2(y3−y1)+x3(y1−y2)2|Area of △ABC=|(3)(2−2+√3)+(−3)(2−√3−1)+0(1−2)2|
=|3√3−3+3√32|
=−3+6√32 sq. units
Let the point on the x-axis be (x,0).Distance between (x,0) and (7,6)=√(7−x)2+(6−0)2=√72+x2−14x+36=√x2−14x+85Distance between (x,0) and (−3,4)=√(−3−x)2+(4−0)2=√9+x2+6x+16=√x2+6x+25As the point (x,0) is equidistant from the two points, both the distances calculated are equal. √x2−14x+85=√x2+6x+25⇒x2−14x+85=x2+6x+25⇒85−25=6x+14x⇒60=20x
⇒x=3Thus, the point is (3,0).
or, PA2=PB2=>(x−3)2+(y−4)2=(x−5)2+(y+2)2x2−6x+9+y2−8y+16=x2−10x+25+y2+4y+44x−12y=4x−3y=1 .......(i)Also Area of PAB=10|(x(4+2)+3(−2−y)+5(y−4)2=10|6x−6−2y+5y−202|=106x+3y−262=±10$$ 6x+2y26=\pm 20 $$$$ 6x+2y =46 ........(ii)or 6x + 2y = 6 .......(iii)Solving equation (i) and (ii) x = 7, y = 2 $$ and solving equation (i) and (iii) x=1,y=0
So, the co-ordinates of P are (7,2)or(1,0)
=>√(−3−x)2+(0−y)2=4
But x=0=>√9+y2=4=>9+y2=16
y2=7
=>y=√7 or −√7
Using the section formula, if a point (x,y) divides theline joining the points (x1,y1) and (x2,y2)externally in the ratio m:n, then (x,y)=(mx2−nx1m−n,my2−ny1m−n)Since, AC=2BC=>ACBC=21Substituting (x1,y1)=(−3,4) and (x2,y2)=(2,1) and m=2,n=1 in the section formula, we get C=(2(2)−1(−3)2−1,2(1)−1(4)2−1)=(7,−2)
Distance between the points (0,3) and (5,0)=√(5−0)2+(0−3)2=√25+9=√34
Let the point on the x-axis be (x,0) Distance between (x,0) and (5,4)=√(5−x)2+(4−0)2=√52+x2−10x+16=√x2−10x+41Distance between (x,0) and (−2,3)=√(−2−x)2+(3−0)2=√22+x2+4x+9=√x2+4x+13
As the point (x,0) is equidistant from the two points, both the distances calculated are equal.
√x2−10x+41=√x2+4x+13
=>x2−10x+41=x2+4x+13
41−13=10x+4x
28=14xx=2. Thus, the point is (2,0).
Using this formula, mid point of AB =D=(2−42,2−42)=(−1,−1)
Distance between two points (x1,y1) and (x2,y2) can be calculated using the formula √(x2−x1)2+(y2−y1)2Distance between the points C(5,−8) and D (−1,−1)=√(−1−5)2+(−1+8)2=√36+49=√85
Distance between two points (x1,y1) and (x2,y2) can be calculatedusing the formula √(x2−x1)2+(y2−y1)2Distance between the points (x+y,x−y) and (x−y,x+y)=√(x−y−x−y)2+(x+y−x+y)2=√(2y)2+(2y)2=2√2y
Since, AD=3AB=>ADBD=32
Using the section formula, if a point (x,y) divides theline joining the points (x1,y1) and (x2,y2)externally in the ratio m:n, then (x,y)=(mx2−nx1m−n,my2−ny1m−n)Substituting (x1,y1)=(1,1) and (x2,y2)=(2,−3) and m=3,n=2 in the section formula, we get C=(3(2)−2(1)3−2,3(−3)−2(1)3−2)=(4,−11)
Distance between two points (x1,y1) and(x2,y2) can be calculated using theformula √(x2−x1)2+(y2−y1)2Distance between the points (3,a) and (4,1)=√10
=>√(4−3)2+(1−a)2=√10
√1+(1−a)2=√10
1+(1−a)2=10
(1−a)2=9
1−a=3 or −3
=>a=−2 or 4
Let A(2,5),B(3,−4) and C(7,10)Distance between two points (x1,y1) and (x2,y2) can be calculated using the formula √(x2−x1)2+(y2−y1)2
Distance between the points A and B AB=√(3+2)2+(−4−5)2=√25+81=√106
Distance between the points B and C BC=√(7−3)2+(10+4)2=√16+196=√212
Distance between the points A and C AC=√(7+2)2+(10−5)2=√81+25=√106
Since, (√212)2=(√106)2+(√106)2, the triangle is a right angled triangle.
Distancebetween two points (x1,y1) and (x2,y2) can be calculated using the formula √(x2−x1)2+(y2−y1)2Distance between (a,2) and (3,4)=8 √(3−a)2+(4−2)2=8√9+a2−6a+4=8√a2−6a+13=8a2−6a+13=64a2−6a−51=0
a=−(−6)±√(−62)−4(1)(−51)2(1)
Consider given, distance between the points (3,k) and (4,1) is √10.
Now,
√(3−4)2+(k−1)2=√10
(−1)2+(k−1)2=10
(k−1)2=9
k−1=±3
If,
k−1=−3
k=−2
if,
k−1=+3
k=4
Hence, this is the answer.
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