Question

find the image of the point (3, 8) with respect to the line X + 3y= 7 ​

Answer 1

Step-by-step explanation:

equation of the given line is

x+3y=7-----(1)

Let a point B(a, b) be the image of the point A(3, 8).

It is given that equation (1) in the perpendicular bisector of AB.

Hence the slope of AB = y2−y1x2−x1

i.e., m1=b−8a−3

Slope of the line (1) is −(coefficientofx)(coefficientofy)

(i.e.,) m2=−13

Since line (1) is perpendicular to AB, the product of their slopes is -1.

(i.e) m1m2=−1

⇒b−8a−3×−13=−1

On simplifying we get,

b−83a−9=1

⇒b−8=3a−9

⇒3a−b=1---------(2)

Mid point of AB = (a+32,b+82)

The midpoint of the line segment AB will also satisfy line (1).

Hence (a+32)+3(b+82)=7

On simplifying we get,

a+3+3b+24=14

⇒a+3b=−13-------(3)

Let us solve equation (2) and (3) for a and b.

3a−b=1

(×3)a+3b=−13

3a−b=1

3a+9b=−39

−10b=40⇒b=−4

∴a=−1

Hence a=−1andb=−4

Thus the image of the given point with respect to the given line is (-1, -4)