Question

Q2. Find the equation of line drawn perpendicular to line
x/4+y/6=1 through the point where it meet the y-axis.​

Answer 1

Answer:

The equation of the given line is 4x+6y=1.

This equation can also be written as  3x+2y−12=0

⇒y=2−3x+6,  which is of the form y=mx+c

∴ slope of given line =−23

∴ slope of line perpendicular to given line  =−{−23}1=32

On substituting x=0 in the given equation of line,

we obtain  6y=1⇒y=6

∴ the given line intersect the y-axis at (0,6)

Hence, the equation of line that has slope 32 and passes through point (0,6) is

(y−6)=32(x−0)

⇒3y−18=2x⇒2x−3y+18=0

Thus, the required equation of the line is 2x−3y+18=0.

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