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Solved Subjective Question on Straight line Set 1

Posted on - 04-01-2017

Math

IIT JEE

Example 1

If the coordinates of the mid-points of the sides of a triangle are (1, 2), (0,–1) and (2, -1). Find the coordinates of its vertices. .

Solution:

Let A(x1, y1), B(x2, y2) and C(x3, y3) be the vertices of DABC. Let
D(1, 2), E(0, –1) and F(2, -1) be the mid-points of sides BC, CA and AB respectively
. .

Since D is the mid-point of BC

\ =1 and = 2


&⇒
x2 + x3 =2 and y2 + y3 =4
. . . (i).

Similarly, E and F are the mid-points of CA and AB respectively,

we have

x1 + x3 = 0 and y1 + y3 = –2 . . . (ii).

x1 + x2 =4 and y1 + y2 = -2 . . . (iii).

From (i), (ii) and (iii), we get

x1 + x2 + x3 =3 and y1 + y2 + y3 = 0 . . . (iv).

From (i) and (iv), we get x1 + 2 = 3 and y1 + 4 = 0

\ x1 = 1 and y1 = -4, so, the coordinates of A are (1, -4)

From (ii) and (iv), we get x2 + 0 = 3 and y2 –2 = 0

\ x2 =3 and y2 = 2, so, coordinates of B are (3, 2)

From (iii) and (iv), we get x3 + 4 = 3 and y3 –2 = 0

\ x3 = -1 and y3 = 2, so, coordinates of C are (-1, 2)

Hence, the vertices of the triangle ABC are

A(1, -4), B(3, 2) and C(-1, 2). .

Example 2

Let the opposite vertices of a square be ( 3, 4) and ( 1, -1). Find the coordinates of the remaining vertices. .

Solution:

Let ABCD be a square and let A(3, 4) and C(1, -1) be the given vertices. Let B(x, y) be the unknown vertex. .

Then, AB = BC
&⇒
AB2 = BC2


&⇒
(x – 3)2 + (y – 4)2 = (x –1)2 + (y +1)2


&⇒
4x + 10 y – 23 = 0


&⇒
x = . . . (i)

In triangle ABC, we have AB2 +BC2 = AC2

(x –3)2 + (y –4)2 + (x –1)2 +(y +1)2 = (3 –1)2 +(4 +1)2


&⇒
x2 +y2 - 4x - 3y –1 = 0
. . . (ii) .

Substituting the value of x from (i) into (ii), we get

+ y2 –– 3y – 1 = 0


&⇒
4y2 -12y + 5 = 0


&⇒
(2y –1) (2y –5) = 0
&⇒
y = ,

hence x = 9/2 or x = -1/2 respectively. .

The required points are (9/2, 1/2) and (-1/2, 5/2). .

Example 3

Find the equation of the straight lines passing through (-2, -7) and having intercept of length 3 units between the straight lines 4x + 3y = 12 and
4x + 3y = 3.

Solution:

Distance between the two given parallel lines

= = =

tanq =
&⇒
tanq =

Slope of the parallel lines = = m2

Also tanq = ±
&⇒


&⇒
m1 + .

The slopes are

(1) m1 = -7/24

(2) m1 = ¥(the line is parallel to the y - axis)

The required equations of the lines are 7x + 24y + 182 = 0 and x+2 =0.

Example 4

The equations of the perpendicular bisectors of the sides AB and AC of a triangle ABC are respectively x – y + 5 = 0 and x + 2y = 0. If the
co-ordinates of A are (1, –2), find the equation of BC.

Solution:

Let OE and OF be the perpendicular bisectors of AB and AC respectively. .

Let B be (x1, y1) and C be (x2, y2). E is the mid-point of AB and F is the mid-point of AC. .

E º , F º

Since E and F lie on OE and OF respectively, x1 - y1 + 13 = 0 . . . (1).

and x2 + 2y2 - 3 = 0 . . . (2).

Also, slope of AB = -1 and that of AC = 2, so that

x1 + y1 + 1 = 0 . . . (3).

and 2x2 - y2 - 4 = 0 . . . (4).

Solving these equations, we get the co-ordinates of B and C as

B º (-7, 6) and C º (11/5, 2/5)


&⇒
Equation of BC is 14x + 23y - 40 = 0.

Example 5

Find a, if (a, a2) lies inside the triangle having sides along the lines
2x + 3y = 1, x + 2y – 3 = 0, 6y = 5x – 1.

Solution:

The vertices of the triangle ABC formed by the given lines are

A º (-7, 5), B º (5/4, 7/8)

C º (1/3, 1/9)

Sign of A w.r.t. BC is –ve.

If P lies in-side the DABC, then sign of P will be same as sign of A w.r.t the line BC .


&⇒
5a - 6a2 - 1 < 0
. . . . . (1).

Similarly 2a + 3a2 - 1 > 0 . . . . . (2).

and, a + 2a2 - 3 < 0 . . . . . (3).

Solving, (1), (2) and (3) for a and then taking intersection

we get a ∈ (1/2, 1) È (-3/2, -1) .

 
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