The perpendicular from the origin to the line | vedclass

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(A) The given equation of the line is $y = mx + c$. The perpendicular from the origin $(0, 0)$ meets the line at the point $P(-1, 2)$. The slope of th

Vedclass · Accepted Answer

$m = \frac{1}{2}, c = \frac{5}{2}$

Vedclass · Answer

(A) The given equation of the line is $y = mx + c$. The perpendicular from the origin $(0, 0)$ meets the line at the point $P(-1, 2)$. The slope of the line segment joining $(0, 0)$ and $(-1, 2)$ is $m_1 = \frac{2 - 0}{-1 - 0} = -2$. Since the line $y = mx + c$ is perpendicular to this segment,the product of their slopes must be $-1$. Therefore,$m 	imes (-2) = -1$,which gives $m = \frac{1}{2}$. Since the point $(-1, 2)$ lies on the line $y = mx + c$,it must satisfy the equation: $2 = m(-1) + c$ Substituting $m = \frac{1}{2}$: $2 = \frac{1}{2}(-1) + c$ $2 = -\frac{1}{2} + c$ $c = 2 + \frac{1}{2} = \frac{5}{2}$. Thus,the values are $m = \frac{1}{2}$ and $c = \frac{5}{2}$.

The perpendicular from the origin to the line $y=mx+c$ meets it at the point $(-1, 2)$. Find the values of $m$ and $c$.

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