Question:
Suppose \(P\) and \(Q\) lie on
\[
3x+4y-4=0
\]
and
\[
5x-y-4=0
\]
respectively. If the midpoint of \(PQ\) is \((1,5)\), then the slope of the line passing through \(P\) and \(Q\) is
Suppose \(P\) and \(Q\) lie on
\[
3x+4y-4=0
\]
and
\[
5x-y-4=0
\]
respectively. If the midpoint of \(PQ\) is \((1,5)\), then the slope of the line passing through \(P\) and \(Q\) is
Show Hint
When midpoint is given, write the second point in terms of the first point using midpoint formula, then substitute into the given line equations.
Updated On: Jun 22, 2026
- \(\frac{83}{35}\)
- \(\frac{63}{35}\)
- \(-\frac34\)
- \(\frac34\)
Show Solution
Verified By Collegedunia
The Correct Option is A
Solution and Explanation
Step 1: Let coordinates of \(P\) and \(Q\).
Let \[ P=(x_1,y_1) \] and \[ Q=(x_2,y_2) \] Since midpoint is \((1,5)\), \[ \frac{x_1+x_2}{2}=1 \] and \[ \frac{y_1+y_2}{2}=5 \] So, \[ x_1+x_2=2 \] and \[ y_1+y_2=10 \]
Step 2: Express \(Q\) in terms of \(P\).
\[ x_2=2-x_1 \] and \[ y_2=10-y_1 \]
Step 3: Use line equations.
Since \(P\) lies on \[ 3x+4y-4=0, \] we get \[ 3x_1+4y_1=4 \] Since \(Q\) lies on \[ 5x-y-4=0, \] we get \[ 5x_2-y_2=4 \] Substitute \(x_2=2-x_1\) and \(y_2=10-y_1\): \[ 5(2-x_1)-(10-y_1)=4 \] \[ 10-5x_1-10+y_1=4 \] \[ y_1-5x_1=4 \]
Step 4: Solve for \(P\).
We have \[ 3x_1+4y_1=4 \] and \[ y_1-5x_1=4 \] From second equation, \[ y_1=5x_1+4 \] Substitute in first equation: \[ 3x_1+4(5x_1+4)=4 \] \[ 3x_1+20x_1+16=4 \] \[ 23x_1=-12 \] \[ x_1=-\frac{12}{23} \] Then, \[ y_1=5\left(-\frac{12}{23}\right)+4 \] \[ y_1=\frac{32}{23} \]
Step 5: Find \(Q\).
\[ x_2=2-x_1=2+\frac{12}{23} \] \[ x_2=\frac{58}{23} \] \[ y_2=10-y_1=10-\frac{32}{23} \] \[ y_2=\frac{198}{23} \]
Step 6: Find slope of \(PQ\).
\[ m=\frac{y_2-y_1}{x_2-x_1} \] \[ m= \frac{\frac{198}{23}-\frac{32}{23}}{\frac{58}{23}+\frac{12}{23}} \] \[ m=\frac{\frac{166}{23}}{\frac{70}{23}} \] \[ m=\frac{166}{70} \] \[ m=\frac{83}{35} \] Therefore, \[ \boxed{\frac{83}{35}} \]
Let \[ P=(x_1,y_1) \] and \[ Q=(x_2,y_2) \] Since midpoint is \((1,5)\), \[ \frac{x_1+x_2}{2}=1 \] and \[ \frac{y_1+y_2}{2}=5 \] So, \[ x_1+x_2=2 \] and \[ y_1+y_2=10 \]
Step 2: Express \(Q\) in terms of \(P\).
\[ x_2=2-x_1 \] and \[ y_2=10-y_1 \]
Step 3: Use line equations.
Since \(P\) lies on \[ 3x+4y-4=0, \] we get \[ 3x_1+4y_1=4 \] Since \(Q\) lies on \[ 5x-y-4=0, \] we get \[ 5x_2-y_2=4 \] Substitute \(x_2=2-x_1\) and \(y_2=10-y_1\): \[ 5(2-x_1)-(10-y_1)=4 \] \[ 10-5x_1-10+y_1=4 \] \[ y_1-5x_1=4 \]
Step 4: Solve for \(P\).
We have \[ 3x_1+4y_1=4 \] and \[ y_1-5x_1=4 \] From second equation, \[ y_1=5x_1+4 \] Substitute in first equation: \[ 3x_1+4(5x_1+4)=4 \] \[ 3x_1+20x_1+16=4 \] \[ 23x_1=-12 \] \[ x_1=-\frac{12}{23} \] Then, \[ y_1=5\left(-\frac{12}{23}\right)+4 \] \[ y_1=\frac{32}{23} \]
Step 5: Find \(Q\).
\[ x_2=2-x_1=2+\frac{12}{23} \] \[ x_2=\frac{58}{23} \] \[ y_2=10-y_1=10-\frac{32}{23} \] \[ y_2=\frac{198}{23} \]
Step 6: Find slope of \(PQ\).
\[ m=\frac{y_2-y_1}{x_2-x_1} \] \[ m= \frac{\frac{198}{23}-\frac{32}{23}}{\frac{58}{23}+\frac{12}{23}} \] \[ m=\frac{\frac{166}{23}}{\frac{70}{23}} \] \[ m=\frac{166}{70} \] \[ m=\frac{83}{35} \] Therefore, \[ \boxed{\frac{83}{35}} \]
Was this answer helpful?
0
0
Top AP EAPCET Mathematics Questions
- $D = \left\{x\in\mathbb{R}: f\left(x\right) =\sqrt{\frac{x - \left|x\right|}{x - \left[x\right]}} \text{is defined} \right\}$ and $C$ be the range of the real function $g(x) = \frac{2x}{4 + x^{2}}$. Then $D \cap C$
- $f\left(x\right) = \frac{x}{e^{x}-1} + \frac{x}{2} + 2 \cos^{3} \frac{x}{2} $ on $R-\left\{0\right\} $ is
- Consider the following lists.
List-I List-II (A) $f(x) = \frac{|x+2|}{x+2} , x \ne -2 $ (I) $[\frac{1}{3} , 1 ]$ (B) $(x)=|[x]|,x \in [R$ (II) Z (C) $h(x) = |x - [x]| , x \in [R$ (III) W (D) $f(x) = \frac{1}{2 - \sin 3x} , x \in [R$ (IV) [0, 1) (V) { -1, 1} - If $[x]$ is the greatest integer less than or equal to $x$ and $|x|$ is the modulus of $x$. then the system of three equations $2x + 3 | y | + 5[z] = 0, x + |y| - 2[z] = 4, x + |y| + |z| = 1$ has
- Investigate the values of $\lambda$ and $\mu$ for the system $x+2 y+3 z=6, x+3 y+5 z=9$, $2 x+5 y+\lambda z=\mu$ and match the values in List - I with the items in List - II.
List I List II (A) $\lambda=8, \mu \neq 15$ 1. Infinitely many solutions (B) $\lambda \neq 8, \mu \in R$ 2. No solution (C) $\lambda=8, \mu=15$ 3. Unique solution
View More Questions
Top AP EAPCET Straight lines Questions
- Find the slope of the line perpendicular to the line $ 3x + 4y - 12 = 0 $.
- The equation of the line passing through the point \((1, 2)\) and perpendicular to the line \(3x + 4y - 12 = 0\) is:
- If the straight lines $x + 2y - 9 = 0$, $3x + 5y - 5 = 0$ and $ax + by - 1 = 0$ are concurrent, then the straight line $22x - 35y = 1$ passes through the point:
- The distance between the parallel lines $5x + 12y - 3 = 0$ and $5x + 12y + 10 = 0$ is:
- The foot of the perpendicular from the point $(1, 3)$ to the line $x + y - 2 = 0$ is:
View More Questions
Top AP EAPCET Questions
- A body is projected from the ground at an angle of $\tan^{-1} \left( \frac{8}{7}\right)$ with the horizontal. The ratio of the maximum height attained by it to its range is
- The four quantum numbers of the valence electron of potassium are :
- A lens forms real and virtual images of an object, when the object is at $u_{1}$ and $u_{2}$ distances respectively. If the size of the virtual image is double that of the real image, then the focal length of the lens is (take, the magnification of the real image as $m$ )
- Two spheres $P$ and $Q$, each of mass $200\, g$ are attached to a string of length one metre as shown in the figure. The string and the spheres are then whirled in a horizontal circle about $O$ at a constant angular speed. The ratio of the tension in the string between $P$ and $Q$ to that of between $P$ and $O$ is $(P$ is at mid-point of the line joining $O$ and $Q$ )
- The volume of $0.02\, M$ acidified permanganate solution required for complete reaction of $60\, mL$ of $0.01\, MI^{-}$ ion solution to form $I_2$ in $mL$ is
View More Questions