   Question Answers

# Find the equation of the line passing through the point (- 4,- 3) and perpendicular to the straight line joining (1,3) and (2,7).  Verified
163.5k+ views Now, we have to find the equation of line passing through point P (- 4,- 3) and perpendicular to the line joining the given points.

First we find the slope of the line formed after joining two points Q (1,3) and R (2,7).

Let the slope of line joining points Q (1,3) and R (2,7) be $m_1$.

Then $m_1 = \dfrac{{{y_2} - {y_1}}}{{{x_2} - {x_1}}} = \dfrac{{7 - 3}}{{2 - 1}} = 4$

Let the slope of required line be ${m_2}$

As, it is given that the required line is perpendicular to the line joining points Q (1,3) and R (2,7).

So, ${m_1}*{m_2} = - 1$

Putting value of ${m_1}$, we get ${m_2} = \dfrac{{ - 1}}{4}$

Now, finding the equation of line passing through point P and having slope ${m_2}$ using point slope form,

$\left( {y - \left( { - 3} \right)} \right) = \dfrac{{ - 1}}{4}\left( {x - \left( { - 4} \right)} \right){\text{ }}\left( 1 \right)$

Cross - multiplying equation 1 we get,

$4y + 12 = - x - 4$

Solving above equation we get $4y + x + 16 = 0$

Hence, equation of line passing through point P and perpendicular to line joining points Q and R is $4y + x + 16 = 0$

NOTE: - Whenever you come up with a type of problem the first find the slope of lines. As for perpendicular lines ${m_1}*{m_2}$ = - 1 and for parallel lines ${m_1} = {m_2}$. Then use point - slope form to find the equation of line.