Find the equations to the altitudes of the triangle whose angular points are $A\left( {2, - 2} \right),B\left( {1{\text{ }},1} \right),C\left( { - 1,0} \right)$ .
Last updated date: 18th Mar 2023
•
Total views: 305.7k
•
Views today: 6.84k
Answer
305.7k+ views
Hint: In order to solve these type of question, we have to simply find out the slopes $m$ between two points i.e. $AB,AC,BC$ using formula $m = \dfrac{{{y_2} - {y_1}}}{{{x_2} - {x_1}}}$ and then convert them into slopes of the altitudes $AD,BE,CF$ by using ${m_{AD}} = - \dfrac{1}{{{m_{BC}}}}$ , ${m_{BE}} = - \dfrac{1}{{{m_{AC}}}}$ , ${m_{CF}} = - \dfrac{1}{{{m_{AB}}}}$ after that substitute the value of ${m_{AD}},{m_{BE}},{m_{CF}}$ in equation for altitude between two points $y - {y_1} = m\left( {x - {x_1}} \right)$ .
Complete step-by-step answer:
Given points are,
$A\left( {2, - 2} \right),B\left( {1{\text{ }},1} \right),C\left( { - 1,0} \right)$
Now we can find the slope of $A\left( {2, - 2} \right),B\left( {1,1} \right)$ by using the formula,
$m = \dfrac{{{y_2} - {y_1}}}{{{x_2} - {x_1}}}$
${m_{AB}} = {\text{ }}\dfrac{{1 - \left( { - 2} \right)}}{{1 - 2}}$
Or ${m_{AB}} = - 3$
Therefore, using ${m_{CF}} = - \dfrac{1}{{{m_{AB}}}}$
${m_{CF}} = {\text{ }}\dfrac{1}{3}$
Using $y - {y_1} = m\left( {x - {x_1}} \right)$
$y - 0 = \dfrac{1}{3}\left( {x - \left( { - 1} \right)} \right)$
Or $3y = x + 1$
Or $x - 3y + 1 = 0 - - - - - - \left( 1 \right)$
Similarly Slope of $B\left( {1{\text{ }},1} \right),C\left( { - 1,0} \right)$ is
${m_{BC}} = \dfrac{{0 - 1}}{{ - 1 - 1}}$
${m_{BC}} = \dfrac{1}{2}$
Therefore, using ${m_{AD}} = - \dfrac{1}{{{m_{BC}}}}$
${m_{AD}} = - 2$
Using $y - {y_1} = m\left( {x - {x_1}} \right)$
$y - \left( { - 2} \right) = \left( { - 2} \right)\left( {x - 2} \right)$
Or $y + 2 = - 2x + 4$
Or $y + 2x - 2 = 0 - - - - \left( 2 \right)$
Slope of $A\left( {2, - 2} \right),C\left( { - 1,0} \right)$
${m_{AC}} = \dfrac{{0 - \left( { - 2} \right)}}{{ - 1 - 2}}$
Or ${m_{AC}} = - \dfrac{2}{3}$
Therefore, using ${m_{BE}} = - \dfrac{1}{{{m_{AC}}}}$
${m_{BE}} = \dfrac{3}{2}$
Now, using $y - {y_1} = m\left( {x - {x_1}} \right)$
$y - 1 = \dfrac{3}{2}\left( {x - 1} \right)$
Or $\left( {y - 1} \right)2 = 3x - 3$
Or $2y - 2 - 3x - 3 = 0$
Or $2y - 3x + 1 = 0 - - - - - \left( 3 \right)$
Therefore, $\left( 1 \right),\left( 2 \right),\left( 3 \right)$ are the equations of CF , AD , and BE which are altitudes of the given triangle.
Note: Whenever we face these type of question the key concept is that firstly we have to find out the slopes of $AB,AC,BC$ and the convert them into the slopes of altitudes $AD,BE,CF$ and then put them in the equation of altitudes between two points and we will easily get our desired equations.
Complete step-by-step answer:

Given points are,
$A\left( {2, - 2} \right),B\left( {1{\text{ }},1} \right),C\left( { - 1,0} \right)$
Now we can find the slope of $A\left( {2, - 2} \right),B\left( {1,1} \right)$ by using the formula,
$m = \dfrac{{{y_2} - {y_1}}}{{{x_2} - {x_1}}}$
${m_{AB}} = {\text{ }}\dfrac{{1 - \left( { - 2} \right)}}{{1 - 2}}$
Or ${m_{AB}} = - 3$
Therefore, using ${m_{CF}} = - \dfrac{1}{{{m_{AB}}}}$
${m_{CF}} = {\text{ }}\dfrac{1}{3}$
Using $y - {y_1} = m\left( {x - {x_1}} \right)$
$y - 0 = \dfrac{1}{3}\left( {x - \left( { - 1} \right)} \right)$
Or $3y = x + 1$
Or $x - 3y + 1 = 0 - - - - - - \left( 1 \right)$
Similarly Slope of $B\left( {1{\text{ }},1} \right),C\left( { - 1,0} \right)$ is
${m_{BC}} = \dfrac{{0 - 1}}{{ - 1 - 1}}$
${m_{BC}} = \dfrac{1}{2}$
Therefore, using ${m_{AD}} = - \dfrac{1}{{{m_{BC}}}}$
${m_{AD}} = - 2$
Using $y - {y_1} = m\left( {x - {x_1}} \right)$
$y - \left( { - 2} \right) = \left( { - 2} \right)\left( {x - 2} \right)$
Or $y + 2 = - 2x + 4$
Or $y + 2x - 2 = 0 - - - - \left( 2 \right)$
Slope of $A\left( {2, - 2} \right),C\left( { - 1,0} \right)$
${m_{AC}} = \dfrac{{0 - \left( { - 2} \right)}}{{ - 1 - 2}}$
Or ${m_{AC}} = - \dfrac{2}{3}$
Therefore, using ${m_{BE}} = - \dfrac{1}{{{m_{AC}}}}$
${m_{BE}} = \dfrac{3}{2}$
Now, using $y - {y_1} = m\left( {x - {x_1}} \right)$
$y - 1 = \dfrac{3}{2}\left( {x - 1} \right)$
Or $\left( {y - 1} \right)2 = 3x - 3$
Or $2y - 2 - 3x - 3 = 0$
Or $2y - 3x + 1 = 0 - - - - - \left( 3 \right)$
Therefore, $\left( 1 \right),\left( 2 \right),\left( 3 \right)$ are the equations of CF , AD , and BE which are altitudes of the given triangle.
Note: Whenever we face these type of question the key concept is that firstly we have to find out the slopes of $AB,AC,BC$ and the convert them into the slopes of altitudes $AD,BE,CF$ and then put them in the equation of altitudes between two points and we will easily get our desired equations.
Recently Updated Pages
Calculate the entropy change involved in the conversion class 11 chemistry JEE_Main

The law formulated by Dr Nernst is A First law of thermodynamics class 11 chemistry JEE_Main

For the reaction at rm0rm0rmC and normal pressure A class 11 chemistry JEE_Main

An engine operating between rm15rm0rm0rmCand rm2rm5rm0rmC class 11 chemistry JEE_Main

For the reaction rm2Clg to rmCrmlrm2rmg the signs of class 11 chemistry JEE_Main

The enthalpy change for the transition of liquid water class 11 chemistry JEE_Main

Trending doubts
Difference Between Plant Cell and Animal Cell

Write an application to the principal requesting five class 10 english CBSE

Ray optics is valid when characteristic dimensions class 12 physics CBSE

Give 10 examples for herbs , shrubs , climbers , creepers

Write the 6 fundamental rights of India and explain in detail

Write a letter to the principal requesting him to grant class 10 english CBSE

List out three methods of soil conservation

Fill in the blanks A 1 lakh ten thousand B 1 million class 9 maths CBSE

Write a letter to the Principal of your school to plead class 10 english CBSE
