Question
A straight line cuts off an intercept of $$2$$ units on the positive direction of $$x$$-axis and passes through the point $$\left( { - 3,\,5} \right).$$ What is the foot of the perpendicular drawn from the point $$\left( {3,\,3} \right)$$ on this line ?
A.
$$\left( {1,\,3} \right)$$
B.
$$\left( {2,\,0} \right)$$
C.
$$\left( {0,\,2} \right)$$
D.
$$\left( {1,\,1} \right)$$
Answer :
$$\left( {1,\,1} \right)$$
Solution :
The given line passes through $$\left( { - 3,\,5} \right)$$ and $$\left( {2,\,0} \right).$$ Its equation is
$$\eqalign{ & y - {y_1} = \left( {\frac{{{y_2} - {y_1}}}{{{x_2} - {x_1}}}} \right)\left( {x - {x_1}} \right) \cr & \Rightarrow \left( {y - 5} \right) = \left( {\frac{{0 - 5}}{{2 + 3}}} \right)\left( {x + 3} \right) \cr & \Rightarrow y = - x + 2......\left( 1 \right) \cr} $$
Slope $$ = m = -1$$ and slope of perpendicular line $$ = - \frac{1}{m} = 1$$
Equation of this line passing through $$\left( {3,\,3} \right)$$ is :
$$\eqalign{ & \left( {y - 3} \right) = 1\left( {n - 3} \right) \cr & \Rightarrow y = x \cr} $$
From equation $$\left( 1 \right)$$ we get,
$$\eqalign{ & x = - x + 2 \cr & \Rightarrow x = 1{\text{ and }}y = 1 \cr} $$
The given line passes through $$\left( { - 3,\,5} \right)$$ and $$\left( {2,\,0} \right).$$ Its equation is
$$\eqalign{ & y - {y_1} = \left( {\frac{{{y_2} - {y_1}}}{{{x_2} - {x_1}}}} \right)\left( {x - {x_1}} \right) \cr & \Rightarrow \left( {y - 5} \right) = \left( {\frac{{0 - 5}}{{2 + 3}}} \right)\left( {x + 3} \right) \cr & \Rightarrow y = - x + 2......\left( 1 \right) \cr} $$
Slope $$ = m = -1$$ and slope of perpendicular line $$ = - \frac{1}{m} = 1$$
Equation of this line passing through $$\left( {3,\,3} \right)$$ is :
$$\eqalign{ & \left( {y - 3} \right) = 1\left( {n - 3} \right) \cr & \Rightarrow y = x \cr} $$
From equation $$\left( 1 \right)$$ we get,
$$\eqalign{ & x = - x + 2 \cr & \Rightarrow x = 1{\text{ and }}y = 1 \cr} $$