Question

Find the equation of the line joining points (3 8) and (7 5) and show that it passes through the point(15 -1)?

Answer 1

The equation of line is `3x+4y =41` which passes through point `(15,-1).`

Explanation:

Let `(x_1=3 , y_1=8 and x_2=7 , y_2=5)`

The slope of the line is `m= (y_2-y_1)/(x_2-x_1) :.`

`m= (5-8)/(7-3) = -3/4`

The equation of line passing through `(x_1,y_1)` is

`y-y_1=m(x-x_1) or y-8 = -3/4(x-3)` or

`4y-32 = -3x+9 or 3x+4y =41`. If the line passes through

point `(15,-1)` , the point will satisfy the equation of line.

Putting `x=15 and y=-1` in the equation of line we get

`:. 3x+4y =41 or 3*15 + 4*(-1 )= 45-4=41 :.`

LHS=RHS , hence the line passes through point `(15,-1)`. {Ans]