Question

Section -C
Case Based MCQ's
Q41).
A parking lot in a company is triangular shaped. Its sides are given by the
equations AB: 3 y = 5x + 2. BC: x+y-6=0 and AC : By-x+2= 0)
Based on the above information answer the following questions:
The coordinates of verte A are
(..) (-1,-1) (b)(-1,2) (c)(1,2) (d)(-1,1)
(i) the coordinates of vertex Bare
(1) (-22) (b)(2,-2) (c)(2, 4) (d)(2,-4)
(ii) Equations of line passing through A and perpendicular to BC is
a) x+y=0 (b) x-y =0) (c) x+2y = 0 (d)x – 2y = 0
iv) Equation of line passing through B and perpendicular to AC is
(a) x+ 3y+10=0. (b) x-3y+10 = 0) (c) 3x -y-10=0 (d) 3 x + y - 10 = 0
) The coordinates of the Orthocentre of triangle ABC are
(-5/2,-5/2) (b)(-5/2,5/2) (c) ( 5/2, 5/2) (d) (5/2,-5/2)

Answer 1

Step-by-step explanation:

Given: A parking lot in a company is triangular shaped. Its sides are given by the equations

AB: 3 y = 5x + 2

BC: x+y-6=0 and AC : 3y-x+2= 0

To find:

Based on the above information answer the following questions:

i) The coordinates of vertex A are

(a) (-1,-1) (b)(-1,2) (c)(1,2) (d)(-1,1)

(ii) the coordinates of vertex B are

(1) (-22) (b)(2,-2) (c)(2, 4) (d)(2,-4)

(iii) Equations of line passing through A and perpendicular to BC is

a) x+y=0 (b) x-y =0) (c) x+2y = 0 (d)x – 2y = 0

iv) Equation of line passing through B and perpendicular to AC is

(a) x+ 3y+10=0. (b) x-3y+10 = 0) (c) 3x -y-10=0 (d) 3 x + y - 10 = 0

v) The coordinates of the Ortho centre of triangle ABC are

(-5/2,-5/2) (b)(-5/2,5/2) (c) ( 5/2, 5/2) (d) (5/2,-5/2)

Solution:

Find at least two points of each line.

Put value of x and find value of y or vice versa

For AB: 3 y = 5x + 2

put x=2

3y=5(2)+2

3y=10+2

3y=12

y=4

One point is (2,4)

like this way find another point, it is (8,14)

Draw these points and draw line AB.

For CB: x+y-6=0

Put x=0

y=6

Point is (0,6)

another point is (6,0)

Draw these points and draw line CB.

For AC: 3y-x+2=0

Put y=0

x=2

Point is (2,0)

another point is (8,2)

Draw these points and draw line AC.

Graph is attatched.

The lines AB, CB and AC meet at three points, these are vertices of triangle ABC.

With the help of graph locate these points. As shown in attachment.

i) The coordinates of vertex A are

(a) (-1,-1) (b)(-1,2) (c)(1,2) (d)(-1,1)

Ans: It is clear from graph that coordinates of A are (-1,-1).

Option (a) is correct.

(ii) the coordinates of vertex B are

(a) (-22) (b)(2,-2) (c)(2, 4) (d)(2,-4)

Ans: It is clear from graph that coordinates of B are (2,4).

Option (c) is correct.

(iii) Equations of line passing through A and perpendicular to BC is

a) x+y=0 (b) x-y =0) (c) x+2y = 0 (d)x – 2y = 0

Ans:

Equation that passes from a point $(x_1,y_1)$ having slope m is given by

$\boxed{\red{y-y_1=m(x-x_1)}}$

We have point A(-1,-1).

Now find slope: As slope of two perpendicular lines must satisfy $\boxed{\bold{\pink{m_1m_2=-1}}}$

If slope of BC can be calculated then slope of perpendicular line can also be calculated.

Slope of line passing through two points $(x_1,y_1)$ and $(x_2,y_2)$ is given by

$\boxed{\green{m = \frac{y_2 - y_1}{x_2 - x_1} }}$

let slope of BC is m1

B(2,4) and C(5,1)

$m_1 = \frac{1 - 4}{5 - 2}$

$m_1 = - \frac{3}{3}$

$m_1 = - 1$

Let slope of line perpendicular to BC is m2

$m_1m_2=-1$

$(-1)m_2=-1$

$m_2=1$

Now find the equation of line which passes from A and perpendicular to BC

$y - ( -1 ) = 1(x - (1))$

$y + 1 = x + 1$

$y - x = 0$

or

$\bold{\red{x- y = 0}}$

Thus,

Option (b) is correct.

iv) Equation of line passing through B and perpendicular to AC is

(a) x+ 3y+10=0. (b) x-3y+10 = 0) (c) 3x -y-10=0 (d) 3 x + y - 10 = 0

Ans:

As discussed in part (iii), find the equation of line which passes through B(2,4) and perpendicular to AC.

let slope of AC is m1

A(-1,-1) and C(5,1)

$m_1 = \frac{1 +1}{5 +1}$

$m_1 = \frac{2}{6}$

$m_1 = \frac{1}{3}$

Let slope of line perpendicular to AC is m2

$m_1m_2=-1$

$\frac{1}{3}m_2=-1$

$m_2=-3$

Now find the equation of line which passes from B and perpendicular to AC

$y - 4= -3(x - 2)$

$y -4 = -3x +6$

$\bold{\green{3x+y-10= 0}}$

Thus,

Option d is correct.

v) The coordinates of the Ortho centre of triangle ABC are

(-5/2,-5/2) (b)(-5/2,5/2) (c) ( 5/2, 5/2) (d) (5/2,-5/2)

Ans:

Both perpendicular intersect each other at a point, which is known as Ortho centre of triangle.

Like AB, CB and AC are drawn, draw the lines calculated in part (iii) and (iv) on the same graph.

Intersection of these are Ortho centre. Coordinates are (5/2,5/2) or (2.5,2.5)

Thus, option c is correct.

Algebraic method: Solve these two equations and find the value of x and y.

Final answer:

(i) Option (a) is correct.

(ii) Option (c) is correct.

(iii) Option (b) is correct.

(iv) Option (d) is correct.

(v) Option (c) is correct.

Hope it helps you.

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Answer 2

Answer:

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Step-by-step explanation:

final answer:-

a,c,b,d,c