Question

Find the coordinates of the point in which $2y-3x+7=0$ meets the line joining the two points $\left( 6,-2 \right)$and $\left( -8,7 \right)$. Find also the angle between them.

Answer 1

Hint: Here, we have to first find the equation of the line joining given points and then solve the 2 equations in two variables simultaneously to find the point of intersection.

Complete step-by-step answer:
Equation of the line joining $\left( 6,-2 \right)$and $\left( -8,7 \right)$is given by
$y-\left( -2 \right)=\dfrac{-7-\left( -2 \right)}{-8-6}=\left( x-6 \right)$
$y+2=\dfrac{-9}{14}\left( x-6 \right)$
$14y+28=-9x+54$
$14y+9x=26\text{ }.........\text{ 1}$
Slope of line will be $y=\dfrac{-9}{14}x+\dfrac{26}{14}$
Comparing with $y=mx+c$we get,
$m=\dfrac{-9}{14}$
${{m}_{1}}=\text{slope}=\dfrac{-9}{14}$
Given line $2y-3x+7=0\text{ }y=\dfrac{3}{2}x-\dfrac{7}{2}$
Slope ${{m}_{2}}=\dfrac{3}{2}$
$\Rightarrow 3x=2y+7\text{ }.........\text{ 2}$
.Substituting 2 into 1 we get,
$14y+3\left( 2y+7 \right)=26$
$\Rightarrow 14y+6y+21=26$
$\Rightarrow 20y=5$
$\Rightarrow y=\dfrac{1}{4}$ Now substituting value of x in 1 we get,
$3x=2\left( \dfrac{1}{4} \right)+7$
$\Rightarrow 3x=\dfrac{15}{2}$
$\Rightarrow x=\dfrac{5}{2}$
So the point of intersection is $\left( \dfrac{5}{2},\dfrac{1}{4} \right)$
Angle between the two lines i.e. $\tan \theta =\left| \dfrac{{{m}_{1}}-{{m}_{2}}}{1+{{m}_{1}}{{m}_{2}}} \right|$
Here ${{m}_{1}}=\dfrac{-9}{14}\text{ }{{m}_{2}}=\dfrac{3}{2}$
$\tan \theta \left| \dfrac{\dfrac{-9}{14}-\dfrac{3}{2}}{1+\left( \dfrac{-9}{14} \right)\left( \dfrac{3}{2} \right)} \right|$
$\tan \theta \left| \dfrac{\dfrac{-18+\left( -42 \right)}{28}}{\dfrac{28-27}{28}} \right|=\left| \dfrac{\dfrac{-60}{28}}{\dfrac{1}{28}} \right|$
$\Rightarrow \tan \theta =60$
$\theta ={{\tan }^{-1}}60$

Note: The equation of line joining two points $\left( {{x}_{1}},{{y}_{1}} \right)\text{ }\\!\\!\And\\!\\!\text{ }\left( {{x}_{2}},{{y}_{2}} \right)$is given by
$y-{{y}_{1}}=\left( \dfrac{{{y}_{2}}-{{y}_{1}}}{{{x}_{2}}-{{x}_{1}}} \right)\left( {{x}_{2}}-{{x}_{1}} \right)$