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).
Accepted Solution
A:
Answer:[tex]y = \frac{3}{4} x+ \frac{7}{4}[/tex] 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 = [tex]\frac{3-(-1)}{2-5} =-\frac{4}{3}[/tex]
Therefore, the slope of the altitude passing through Q(-1,1) will be [tex]\frac{3}{4}[/tex] {Since, the product of slopes of two perpendicular straight line is -1}
So, equation of the altitude is [tex]y=\frac{3}{4} x + c[/tex] where c is a constant.
Now, putting x = -1 and y = 1 in the above equation we get [tex]1 = -\frac{3}{4} + c[/tex]
⇒ [tex]c=\frac{7}{4}[/tex]
Therefore, the equation of the altitude is [tex]y = \frac{3}{4} x+ \frac{7}{4}[/tex] (Answer)