The point P(a, b) lies on the straight line 3 | vedclass

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

Question

(B) Given that point $P(a, b)$ lies on $3x + 2y = 13$,we have: $3a + 2b = 13$ ..... $(i)$ Given that point $Q(b, a)$ lies on $4x - y = 5$,we have: $4b

Vedclass · Accepted Answer

$x + y = 5$

Vedclass · Answer

(B) Given that point $P(a, b)$ lies on $3x + 2y = 13$,we have: $3a + 2b = 13$ ..... $(i)$ Given that point $Q(b, a)$ lies on $4x - y = 5$,we have: $4b - a = 5$ or $-a + 4b = 5$ ..... $(ii)$ Multiplying equation $(ii)$ by $3$,we get $-3a + 12b = 15$ ..... $(iii)$ Adding $(i)$ and $(iii)$: $(3a + 2b) + (-3a + 12b) = 13 + 15$ $14b = 28 \Rightarrow b = 2$ Substituting $b = 2$ in $(i)$: $3a + 2(2) = 13$ $\Rightarrow 3a = 9$ $\Rightarrow a = 3$ Thus,$P = (3, 2)$ and $Q = (2, 3)$. The slope $m$ of line $PQ$ is $\frac{3 - 2}{2 - 3} = \frac{1}{-1} = -1$. The equation of line $PQ$ passing through $(3, 2)$ is: $y - 2 = -1(x - 3)$ $y - 2 = -x + 3$ $x + y = 5$.

The point $P(a, b)$ lies on the straight line $3x + 2y = 13$ and the point $Q(b, a)$ lies on the straight line $4x - y = 5$. Then,the equation of the line $PQ$ is:

Similar Questions

The number of lines passing through $A(3, 4)$ such that the sum of their non-zero intercepts is zero is:

The polar equation of the line perpendicular to the line $\sin \theta - \cos \theta = \frac{1}{r}$ and passing through the point $\left(2, \frac{\pi}{6}\right)$ is

The line passing through $\left(-1, \frac{\pi}{2}\right)$ and perpendicular to $\sqrt{3} \sin \theta + 2 \cos \theta = \frac{4}{r}$ is:

Let a line intersect the co-ordinate axes in points $A$ and $B$ such that the area of the triangle $OAB$ is $12$ sq. units. If the line passes through the point $(2,3)$,then the equation of the line is

The equation of the line passing through $(1, 1)$ and parallel to the line $2x + 3y - 7 = 0$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module