The point on the line 2x-3y=5 that is equidistant from (1,2) and (3,4) is (1) (7,3) (2) (4,1) (3) (1,-1) (4) (

Exercise 6.5 · Q15

2x-3y=5

(1)

(7,3)

(2)

(4,1)

(3)

(1,-1)

(4)

(-2,3)

Tamil Nadu DgeTextbookSubjectiveImportance★★★★★

64% · 83/129 Questions

🔒 Locked · start free trial →

You're viewing a preview — the full solution, concept, methods & PYQ mapping are locked.

Start your 14-day free trial to unlock the full solution →

Concept understanding — Distance and Concurrency

Three distance formulas underpin this chapter's metric questions:

Point to point. $D=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}$ (the ordinary Pythagorean distance).
Point to line. The distance from $P(x_1,y_1)$ to the line $ax+by+c=0$ is

D=\left|\dfrac{ax_1+by_1+c}{\sqrt{a^2+b^2}}\right|,

proved by comparing the normal form of the given line with the normal form of the parallel line through $P$ , and taking the difference of their normal lengths from the origin.

3. Line to parallel line. For $a_1x+b_1y+c_1=0$ and $a_1x+b_1y+c_2=0$ (same $a_1,b_1$ ), $D=\dfrac{|c_2-c_1|}{\sqrt{a_1^2+b_1^2}}$ — a special case of formula 2, taking the point as the origin (or any point) on one of the two lines.

Foot of the perpendicular and image of a point. From $P(x_1,y_1)$ , the foot of the perpendicular on $ax+by+c=0$ is found from the parametric relation

\dfrac{x-x_1}{a}=\dfrac{y-y_1}{b}=\dfrac{-(ax_1+by_1+c)}{a^2+b^2},

and the image (reflection) of $P$ in that line doubles the same offset:

\dfrac{x-x_1}{a}=\dfrac{y-y_1}{b}=\dfrac{-2(ax_1+by_1+c)}{a^2+b^2}.

(The foot is the midpoint of $P$ and its image.) The same idea — reflect a point in a line, most often the $x$ -axis or $y$ -axis — is the standard tool for 'ray of light reflected off a surface' problems: the angle of incidence equals the angle of reflection exactly when the reflected ray is the straight line joining the image of the source to the target point.

Position of a point relative to a line and the acute-angle test. A point $P(x_1,y_1)$ lies on the origin side or non-origin side of $ax+by+c=0$ ( $c\ne0$ ) according as $ax_1+by_1+c$ has the same or opposite sign as $c$ . …

Start your 14-day free trial to unlock the full solution →

A point equidistant from $(1,2)$ and $(3,4)$ lies on their perpendicular bisector $x+y=5$ ; intersect it with $2x-3y=5$ .

Any point equidistant from two given points must lie on the perpendicular bisector of the segment joining them. So the required point is the intersection of the given line with that perpendicular bisector.

Step 1. Find the midpoint of $(1,2)$ and $(3,4)$ .

M=\left(\dfrac{1+3}{2},\dfrac{2+4}{2}\right)=(2,3)

Step 2. Find the slope of the segment and of its perpendicular. Slope of the segment joining $(1,2)$ and $(3,4)$ :

m=\dfrac{4-2}{3-1}=1

The perpendicular bisector has slope $-\dfrac{1}{m}=-1$ .

Step 3. Write the equation of the perpendicular bisector through $M(2,3)$ with slope $-1$ :

y-3=-1(x-2) \implies y=-x+5 \implies x+y=5

Step 4. Intersect with the given line $2x-3y=5$ . From $x+y=5$ , $x=5-y$ . Substitute: …

Start your 14-day free trial to unlock the full solution →

←Previous question · 6.5.4Q14 — The line (p+2q)x+(p-3q)y=p-q, for different values of p and q, passes …→Next question · 6.5.4Q16 — The image of the point (2, 3) in the line y=-x is (1) (-3,-2) (2) (-3,…

Q.The point on the line 2x−3y=52x-3y=52x−3y=5 that is equidistant from (1,2)(1,2)(1,2) and (3,4)(3,4)(3,4) is

🎓Unlock everything free for 14 days

Q.The point on the line $2x-3y=5$ that is equidistant from $(1,2)$ and $(3,4)$ is

Unlock everything free for 14 days