The vertex A of a triangle lies on the lines | vedclass

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

Question

(B) Vertex $A$ is the point of intersection of the lines $x+y=1$ and $2x+3y=6$. Solving these,we get $x = -3$ and $y = 4$. So,$A(-3, 4)$ and $O\left(\

Vedclass · Accepted Answer

$x \cos \alpha+y \sin \alpha=\frac{13}{\sqrt{17}} ; \alpha=	an ^{-1}(4)$

Vedclass · Answer

(B) Vertex $A$ is the point of intersection of the lines $x+y=1$ and $2x+3y=6$. Solving these,we get $x = -3$ and $y = 4$. So,$A(-3, 4)$ and $O\left(\frac{3}{7}, \frac{22}{7}ight)$. The equation of the line $OA$ passing through $A(-3, 4)$ and $O\left(\frac{3}{7}, \frac{22}{7}ight)$ is: $y - 4 = \frac{\frac{22}{7} - 4}{\frac{3}{7} - (-3)}(x + 3)$ $y - 4 = \frac{-6/7}{24/7}(x + 3)$ $y - 4 = -\frac{1}{4}(x + 3)$ $4y - 16 = -x - 3 \Rightarrow x + 4y = 13$. The normal form of the line $x \cos \alpha + y \sin \alpha = p$ is obtained by dividing by $\sqrt{1^2 + 4^2} = \sqrt{17}$. $\frac{1}{\sqrt{17}}x + \frac{4}{\sqrt{17}}y = \frac{13}{\sqrt{17}}$. Here,$\cos \alpha = \frac{1}{\sqrt{17}}$ and $\sin \alpha = \frac{4}{\sqrt{17}}$,so $	an \alpha = 4$,which means $\alpha = 	an^{-1}(4)$. Thus,the equation is $x \cos \alpha + y \sin \alpha = \frac{13}{\sqrt{17}}$ with $\alpha = 	an^{-1}(4)$.

The vertex $A$ of a triangle lies on the lines $x+y=1$ and $2x+3y=6$. If the orthocentre of the triangle is $O\left(\frac{3}{7}, \frac{22}{7}\right)$,then the equation of $OA$ in the normal form is

Similar Questions

If the midpoints of the sides $BC, CA$ and $AB$ of the triangle $ABC$ are $(1, 3), (5, 7)$ and $(-5, 7)$ respectively,then the equation of the side $AB$ is

The number of integral values of $m$ such that the abscissa of the point of intersection of the lines $3x + 4y = 9$ and $y = mx + 1$ is also an integer,is

The angles made with the $X$-axis by the two lines passing through the point $P(1, 2)$ and cutting the line $x+y=4$ at a distance $\frac{\sqrt{6}}{3}$ units from the point $P$ are

What is the normal form of the line $x + \sqrt{3}y - 4 = 0$?

$A$ line meets the $x$-axis and $y$-axis at the points $A$ and $B$ respectively. If the midpoint of $AB$ is $(x_1, y_1)$,then the equation of the line is:

Vedclass Test Series

Exam Paper Generator

Online Exam Module