The perpendicular from the origin to the line | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(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 $c$.

Similar Questions

Starting from the point $A(-3, 4)$,a moving object touches the line $2x + y - 7 = 0$ at point $B$ and reaches the point $C(0, 1)$. If the object travels along the shortest path,the distance between $A$ and $B$ is:

If $(-2, 6)$ is the image of the point $(4, 2)$ with respect to the line $L = 0$,then $L$ is equal to

$A$ light beam emanating from the point $A(3, 10)$ reflects from the line $2x + y - 6 = 0$ and then passes through the point $B(5, 6)$. The equations of the incident and reflected beams are respectively:

The coordinates of the foot of the perpendicular from the point $(2, 3)$ on the line $x + y - 11 = 0$ are

The image of a point $A(3, 8)$ in the line $x + 3y - 7 = 0$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module