Let A be the point of intersection of the lin | vedclass

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

Question

(D) Step $1$: Find point $A$ by solving $3x + 2y = 14$ and $5x - y = 6$. Multiplying the second equation by $2$ gives $10x - 2y = 12$. Adding this to

Vedclass · Accepted Answer

$6$

Vedclass · Answer

(D) Step $1$: Find point $A$ by solving $3x + 2y = 14$ and $5x - y = 6$. Multiplying the second equation by $2$ gives $10x - 2y = 12$. Adding this to the first equation gives $13x = 26$,so $x = 2$. Substituting $x = 2$ into $5x - y = 6$ gives $10 - y = 6$,so $y = 4$. Thus,$A = (2, 4)$.Step $2$: Find point $B$ by solving $4x + 3y = 8$ and $6x + y = 5$. Multiplying the second equation by $3$ gives $18x + 3y = 15$. Subtracting the first equation from this gives $14x = 7$,so $x = \frac{1}{2}$. Substituting $x = \frac{1}{2}$ into $6x + y = 5$ gives $3 + y = 5$,so $y = 2$. Thus,$B = (\frac{1}{2}, 2)$.Step $3$: Find the equation of line $AB$ passing through $(2, 4)$ and $(\frac{1}{2}, 2)$. The slope $m = \frac{2 - 4}{1/2 - 2} = \frac{-2}{-3/2} = \frac{4}{3}$. The equation is $y - 4 = \frac{4}{3}(x - 2)$,which simplifies to $3y - 12 = 4x - 8$,or $4x - 3y + 4 = 0$.Step $4$: Calculate the distance of $P(5, -2)$ from $4x - 3y + 4 = 0$ using the formula $d = \frac{|ax_0 + by_0 + c|}{\sqrt{a^2 + b^2}}$.$d = \frac{|4(5) - 3(-2) + 4|}{\sqrt{4^2 + (-3)^2}} = \frac{|20 + 6 + 4|}{\sqrt{16 + 9}} = \frac{30}{5} = 6$.

Let $A$ be the point of intersection of the lines $3x + 2y = 14$ and $5x - y = 6$. Let $B$ be the point of intersection of the lines $4x + 3y = 8$ and $6x + y = 5$. The distance of the point $P(5, -2)$ from the line $AB$ is:

Similar Questions

$A$ straight line through the origin $O$ meets the parallel lines $4x + 2y = 9$ and $2x + y + 6 = 0$ at $P$ and $Q$ respectively. The point $O$ divides the segment $PQ$ in the ratio

Find the distance between the lines $3x + 4y + 7 = 0$ and $3x + 4y + 22 = 0$.

The length of the perpendicular from the point $(a \cos \alpha, a \sin \alpha)$ to the line $y = x \tan \alpha + c, c > 0$ is .....

The equation of the line passing through $(1, 2)$ and having a distance equal to $7$ units from the point $(8, 9)$ is:

The distance of the point $(1, 2)$ from the line $x + y + 5 = 0$ measured along the line parallel to $3x - y = 7$ is equal to

Vedclass Test Series

Exam Paper Generator

Online Exam Module