Question

The points P(2,3) , Q(-1,1) and R(5,-1) are the vertices of a triangle PQR.Find the equation of the altitude of the triangle PQR drawn from the point Q(-1,1).​

Answer 1

Answer:

$y = \frac{3}{4} x+ \frac{7}{4}$

Step-by-step explanation:

The slope of straight line PR where P(2,3) and R(5,-1) are two vertices of triangle PQR will be = $\frac{3-(-1)}{2-5} =-\frac{4}{3}$

Therefore, the slope of the altitude passing through Q(-1,1) will be $\frac{3}{4}$ {Since, the product of slopes of two perpendicular straight line is -1}

So, equation of the altitude is $y=\frac{3}{4} x + c$ where c is a constant.

Now, putting x = -1 and y = 1 in the above equation we get  

$1 = -\frac{3}{4} + c$

⇒ $c=\frac{7}{4}$

Therefore, the equation of the altitude is $y = \frac{3}{4} x+ \frac{7}{4}$ (Answer)