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Solved Subjective Question on Straight line Set 2

Posted on - 04-01-2017

Math

IIT JEE

Example 1

Prove that (4, -1), (6, 0), (7,2) and (5, 1) are the vertices of rhombus. Is it a square?.

Solution:

Let the given points be A, B, C and D respectively. .

Then, coordinates of mid-point of AC are

Coordinates of the mid-points of BD are

Thus, AC and BD have the same mid-point. Hence, ABCD is a parallelogram. .

Now, AB =

BC =

\ AB = BC. .

So, ABCD is a parallelogram whose adjacent sides are equal

Hence, ABCD is a rhombus. .

Now, AC = and BD =.

Clearly, AC > BD. So, ABCD is not a square. .

Eaxmple 2

Let P (sinq, cosq) (0 £ q £ 2p) be a point and let OAB be a triangle with vertices (0, 0),and Find q if P lies inside the DOAB.

Solution:

Equations of lines along OA, OB and AB are y = 0, x = 0, and x+y =respectively. Now P and B will lie on the same side of y = 0 if cosq > 0. Similarly P and A will lie on the same side of x = 0 if sin q > 0 and P and O will lie on the same side of x + y = if sin q + cosq < .
Hence P will lie inside the D ABC,

if sinq > 0, cosq > 0 and sinq + cosq <.

Now sinq + cosq <


&⇒
sin ( q + p /4) <

i.e. 0 < q + p/4 < p/3 .

or < q + p/4 < p

Since sinq>0 and cos q > 0, so

0< q<

Example 3

A line intersects the straight lines 5x – y – 4 = 0 and 3x – 4y – 4 = 0 at A and B respectively. If a point P (1, 5) on the line AB is such that
AP: PB = 2 : 1 (internally), find the point A.

Solution:

Let AP = 2r. Then PB = r.

Let the slope of AB be tanq. .

The equation of the line AB is

= r, so that

A º (1 + 2rcosq, 5 + 2rsinq)

and Bº (1 – rcosq, 5 – rsinq)

or Aº (1 - 2rcosq, 5 - 2rsinq)

and Bº (1 + rcosq, 5 + rsinq)

The point ‘A’ lies on the line

5x – y – 4 = 0, So that

5 (1 + 2rcosq) - (5 +2rsinq) - 4 = 0 ….(i).

The Point ‘B’ lies on the line 3x – 4y – 4 0,

So that 3 (1–rcosq) + 4(5–rsinq) – 4 = 0 ….(ii).

Let rcosq = a and rsinq = b. .


&⇒
10a – 2b – 4 = 0 and –3a – 4b + 19 = 0

On solving for a and b, we get

Hence the point A is

Similarly when A º (1 - 2rcosq, 5 - 2rsinq)

and B º (1 + rcosq, 5 + rsinq), we get

- 10a + 2b - 4 = 0 and 3a +4b +19 = 0


&⇒
a = , b =

Hence the point A is .

Example 4

Four points A(a, 0), B(b, 0), C(g, 0) and D(d, 0), with a < b < g < d or
a > b > g > d , are such that a and b are the roots of ax2 +2hx + b = 0 and g and d are the roots of a¢x2 +2h¢x + b¢ = 0. If = l, = m and
ab¢ + ba¢ = 2hh¢, show that l + m =0.

Solution:

We have a + b = -2h/a, ab = b/a. Also, g + d = , gd =


&⇒
l = . Also, m = , l + m = +

= =

= =

= = 0.

Example 5

Find the points farthest and nearest on the curve 5x2 + 5y2 + 6xy– 8 = 0 from origin. Also determine the equations of the lines through the origin on which the points are occurring. .

Solution:

Any line through the origin is Any point on this line is
(r cosq, r sinq). If this point lies on the given curve,

then 5r2 + 6r2sinq cosq = 8


&⇒
r2 =


&⇒
rmax = 2 (for q= ) and rmin = 1 (for q = p/4).

The required points are () and (). The lines are
y = – x and y = x.


&⇒

 
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