CBSE Questions for Class 11 Engineering Maths Straight Lines Quiz 10 - MCQExams.com

We have,

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

$$ B\left( {{x}_{2}},{{y}_{2}} \right)=\left( {{x}_{2}},{{y}_{2}} \right) $$

$$ C\left( {{x}_{3}},{{y}_{3}} \right)=\left( 3{{y}_{2}},-2{{y}_{1}} \right) $$

We know that the area of triangle is

$$ Area\,of\,\Delta ABC=\dfrac{1}{2}\left[ {{x}_{1}}\left( {{y}_{2}}-{{y}_{3}} \right)+{{x}_{2}}\left( {{y}_{3}}-{{y}_{1}} \right)+{{x}_{3}}\left( {{y}_{1}}-{{y}_{2}} \right) \right] $$

$$ =\dfrac{1}{2}\left[ {{x}_{1}}\left( {{y}_{2}}+2{{y}_{1}} \right)+{{x}_{2}}\left( -2{{y}_{1}}-{{y}_{1}} \right)+3{{y}_{2}}\left( {{y}_{1}}-{{y}_{2}} \right) \right] $$

$$ =\dfrac{1}{2}\left[ {{x}_{1}}{{y}_{2}}+2{{x}_{1}}{{y}_{1}}-3{{x}_{2}}{{y}_{1}}+3{{y}_{2}}{{y}_{1}}-3{{y}_{2}}^{2} \right] $$

Hence, the is the answer

Consider the given point

$$ A\left( {{x}_{1}},{{y}_{1}} \right)=\left( 0,0 \right) $$

$$ B\left( {{x}_{2}},{{y}_{2}} \right)=\left( 3,\sqrt{3} \right) $$

$$ C\left( {{x}_{3}},{{y}_{3}} \right)=\left( p,q \right) $$

Then,

We know that it is an equilateral triangle,

$$ AB=BC $$

$$ \Rightarrow \sqrt{{{\left( {{x}_{1}}-{{x}_{2}} \right)}^{2}}+{{\left( {{y}_{1}}-{{y}_{2}} \right)}^{2}}}=\sqrt{{{\left( {{x}_{2}}-{{x}_{3}} \right)}^{2}}+{{\left( {{y}_{2}}-{{y}_{3}} \right)}^{2}}} $$

$$ \Rightarrow \sqrt{{{\left( 0-3 \right)}^{2}}+{{\left( 0-\sqrt{3} \right)}^{2}}}=\sqrt{{{\left( 3-p \right)}^{2}}+{{\left( \sqrt{3}-q \right)}^{2}}} $$

$$ \Rightarrow \sqrt{12}=\sqrt{9+{{p}^{2}}-6p+3+{{q}^{2}}-2\sqrt{3}q} $$

$$ \text{Squaring both side and we get,} $$

$$ \Rightarrow 12={{p}^{2}}+{{q}^{2}}-6p-2\sqrt{3}q+12 $$

$$ \Rightarrow {{p}^{2}}+{{q}^{2}}-6p-2\sqrt{3}q=0$$ ....... (1)

Again,

$$ AB=CA $$

$$ \Rightarrow \sqrt{{{\left( {{x}_{1}}-{{x}_{2}} \right)}^{2}}+{{\left( {{y}_{1}}-{{y}_{2}} \right)}^{2}}}=\sqrt{{{\left( {{x}_{3}}-{{x}_{1}} \right)}^{2}}+{{\left( {{y}_{3}}-{{y}_{1}} \right)}^{2}}} $$

$$ \Rightarrow \sqrt{{{\left( 0-3 \right)}^{2}}+{{\left( 0-\sqrt{3} \right)}^{2}}}=\sqrt{{{\left( p-0 \right)}^{2}}+{{\left( q-0 \right)}^{2}}} $$

$$ \Rightarrow \sqrt{12}=\sqrt{{{p}^{2}}+{{q}^{2}}} $$

$$ \text{squaring both side and we get,} $$

$${{p}^{2}}+{{q}^{2}}=12$$…… (2)

By equation (1) and (2) to, and we get,

$$ 12-6p-2\sqrt{3}q=0 $$

$$ \Rightarrow 6p+2\sqrt{3}q=12 $$

$$ \Rightarrow 3p+\sqrt{3}q=6 $$

$$ \Rightarrow 3p=6-\sqrt{3}q $$

$$ \Rightarrow p=\dfrac{6-\sqrt{3}q}{3} $$

Put the value of p in equation (2) and we get,

$$ {{p}^{2}}+{{q}^{2}}=12 $$

$$ \Rightarrow {{\left( \dfrac{6-\sqrt{3}q}{3} \right)}^{2}}+{{q}^{2}}=12 $$

$$ \Rightarrow 36+3{{q}^{2}}-12\sqrt{3}q+9{{q}^{2}}=108$$

$$ \Rightarrow 12{{q}^{2}}-12\sqrt{3}q=72 $$

$$\begin{align}

$$ \Rightarrow 12\left( {{q}^{2}}-\sqrt{3}q-6 \right)=0 $$

$$ \Rightarrow {{q}^{2}}-\sqrt{3}q-6=0 $$

$$ \Rightarrow {{q}^{2}}-2\sqrt{3}q+\sqrt{3}q-6=0 $$

$$ \Rightarrow q\left( q-2\sqrt{3} \right)+\sqrt{3}\left( q-2\sqrt{3} \right)=0 $$

$$ \Rightarrow \left( q-2\sqrt{3} \right)\left( q+\sqrt{3} \right)=0 $$

Then, $$q=2\sqrt{3},\,\,q=-\sqrt{3}$$

Now, According to given question,

$$ q={{q}_{1}}+{{q}_{2}} $$

$$ q=2\sqrt{3}-\sqrt{3} $$

$$ q=\sqrt{3} $$

Hence, this is the answer.

We have,

Since, $$C$$ trisects $$AB$$

Now,

$$ {{x}_{1}}=1,\,\,{{y}_{1}}=\ln 1=0 $$

$$ {{x}_{2}}=1000,\,\,{{y}_{2}}=\ln 1000=3 $$

Now, Ratio$${{m}_{1}}:{{m}_{2}}=1:2$$

$$ a=\dfrac{{{m}_{1}}{{x}_{2}}+{{m}_{2}}{{x}_{1}}}{{{m}_{1}}+{{m}_{2}}} $$

$$ =\dfrac{1\times 1000+2\times 1}{1+2} $$

$$ =\dfrac{1002}{3} $$

$$ a=334 $$

$$ b=\dfrac{{{m}_{1}}{{y}_{2}}+{{m}_{2}}{{y}_{1}}}{{{m}_{1}}+{{m}_{2}}} $$

$$ =\dfrac{1\times \ln 1000+2\times 0}{1+2} $$

$$ b=\dfrac{\ln 1000}{3}\,\,\,\,.......\,\,\left( 1 \right) $$

Given the curve

$$y=\ln x\,\,\,......\,\,\left( 2 \right)$$

From equation (1) and (2) to,

$$ \ln x=\dfrac{\ln 1000}{3} $$

$$ 3\ln x=\ln 1000 $$

$$ \ln x={{\left( \ln 1000 \right)}^{\dfrac{1}{3}}} $$

$$ x={{\left( 1000 \right)}^{\dfrac{1}{3}}} $$

$$ x=10 $$

Hence,$${{x}_{3}}=10$$

$$\textbf{Step -1: Finding all the side lengths.}$$

                  $$\text{Let }(4,0)\text{ be point }A,\;(-1,-1)\text{ be point }B,\;(3,5)\text{ be point }C$$

                  $$AB=\sqrt{(4-(-1))^2+(0-(-1))^2}=\sqrt{5^2+1^2}=\sqrt{26}$$

                  $$BC=\sqrt{(-1-3)^2+(-1-5)^2}=\sqrt{(-4)^2+(-6)^2}=\sqrt{52}$$

                  $$CA=\sqrt{(3-4)^2+(5-0)^2}=\sqrt{(-1)^2+(5)^2}=\sqrt{26}$$

                  $$\therefore AB=CA\text{ Hence, the triangle is isosceles.}$$

$$\textbf{Step -2: Checking whether the triangle is right angled.}$$

                  $$AB^2=26$$

                  $$BC^2=52$$

                  $$CA^2=26$$

                  $$\Rightarrow AB^2+CA^2=26+26=52=BC^2$$

                  $$\text{Hence, the triangle is right angled.}$$

$$\textbf{Hence, the triangle with vertices (4,0), (-1,-1), (3,5) is isosceles and right angled.}$$