Suppose that the points (h, k),(1, 2),and (-3 | vedclass

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

Question

(B) The slope of line $L_1$ passing through $(1, 2)$ and $(-3, 4)$ is $m_1 = \frac{4 - 2}{-3 - 1} = \frac{2}{-4} = -\frac{1}{2}$.Since $(h, k)$ lies o

Vedclass · Accepted Answer

$\frac{1}{3}$

Vedclass · Answer

(B) The slope of line $L_1$ passing through $(1, 2)$ and $(-3, 4)$ is $m_1 = \frac{4 - 2}{-3 - 1} = \frac{2}{-4} = -\frac{1}{2}$.Since $(h, k)$ lies on $L_1$,the slope of the line segment joining $(h, k)$ and $(1, 2)$ must also be $-\frac{1}{2}$.Thus,$\frac{k - 2}{h - 1} = -\frac{1}{2}$ $\Rightarrow 2k - 4 = -h + 1$ $\Rightarrow h + 2k = 5$.Line $L_2$ passes through $(h, k)$ and $(4, 3)$,so its slope is $m_2 = \frac{3 - k}{4 - h}$.Since $L_2 \perp L_1$,$m_1 	imes m_2 = -1$ $\Rightarrow -\frac{1}{2} 	imes \frac{3 - k}{4 - h} = -1$ $\Rightarrow \frac{3 - k}{4 - h} = 2$.$3 - k = 8 - 2h \Rightarrow 2h - k = 5$.We have the system of equations:$1) h + 2k = 5$$2) 2h - k = 5$Multiplying equation $(2)$ by $2$,we get $4h - 2k = 10$.Adding this to equation $(1)$: $(h + 2k) + (4h - 2k) = 5 + 10$ $\Rightarrow 5h = 15$ $\Rightarrow h = 3$.Substituting $h = 3$ into $2h - k = 5$: $2(3) - k = 5$ $\Rightarrow 6 - k = 5$ $\Rightarrow k = 1$.Therefore,$\frac{k}{h} = \frac{1}{3}$.

Suppose that the points $(h, k)$,$(1, 2)$,and $(-3, 4)$ lie on the line $L_1$. If a line $L_2$ passing through the points $(h, k)$ and $(4, 3)$ is perpendicular to $L_1$,then $\frac{k}{h}$ equals

Similar Questions

$(\sqrt{3}+i)^7+(\sqrt{3}-i)^7$ is equal to (in $\sqrt{3}$)

Two oxides of nitrogen,$NO$ and $NO_2$,react together at $253 \ K$ to form a compound of nitrogen $X$. $X$ reacts with water to yield another compound of nitrogen $Y$. The shape of the anion of $Y$ molecule is:

When $FeCl_3$ reacts with $K_4[Fe(CN)_6]$,which of the following compounds is formed?

In a biprism experiment,if the $5^{\text{th}}$ bright band with wavelength $\lambda_1$ coincides with the $6^{\text{th}}$ dark band with wavelength $\lambda_2$,then the ratio $\left(\frac{\lambda_2}{\lambda_1}\right)$ is:

Vedclass Test Series

Exam Paper Generator

Online Exam Module