Find the equation of a line drawn perpendicular to the line $$\dfrac{x}{4}+\dfrac{y}{6}=1$$through the point, where it meets the y-axis.

Asked by Abhisek | 1 year ago |  76

##### Solution :-

It is given that

$$\dfrac{x}{4}+\dfrac{y}{6}=1$$

We can write it as

3x + 2y – 12 = 0

So we get

y = $$- \dfrac{3}{2}x$$+ 6, which is of the form y = mx + c

Here the slope of the given line = $$- \dfrac{3}{2}x$$

So the slope of line perpendicular to the given line

=  =$$\dfrac{-1}{ - \dfrac{3}{2}}= \dfrac{2}{3}$$

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

By substituting x as zero in the equation of the given line

$$\dfrac{y}{6}$$ = 1

y = 6

Hence, the given line intersects the y-axis at (0, 6)

We know that the equation of the line that has a slope of$$\dfrac{2}{3}$$ and passes through point (0, 6) is

(y – 6) = $$\dfrac{2}{3}$$(x – 0)

By further calculation

3y – 18 = 2x

So we get

2x – 3y + 18 = 0

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

Answered by Pragya Singh | 1 year ago

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