Straight Lines Test 28

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Straight Lines Test 28

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Weekly Quiz Competition

Question 1
1 / -0

If the equation $$ax^{2}-6xy+y^{2} + 2gx+2fy+c=0$$ represents a pair of line whose slopes are $$m$$ and $$m^{2}$$, then sum of all possible values of $$a$$ is-

A
$$17$$

B
$$-19$$

C
$$19$$

D
$$-17$$

Solution

$$ax^{2}-6xy+y^{2}+2gx+2fy+c=0$$
$$=(y+3x+c_{1})(y+3^{2}x+c_{2})$$
$$y^{2}+m^{2}xy+c_{2}y+mxy+m^{3}x^{2}+mxc_{2}+c_{1}y+c_{1}m^{2}x+c_{2}c_{2}$$
$$a= m^{3}$$
$$m^{2}+m=-6\Rightarrow m^{2}+m-6=0$$
$$a=8, -27$$ $$m^{2}+3m-2m-6=0$$
$$m(m+3)-2(m+3)=0$$
$$(m-2)(m+3)=0$$
$$m= 2,-3$$
sum of total possible of a
= -19
Question 2
1 / -0

The diagonal passing through origin of a quadrilateral formed by $$x=0, y=0, x+y=1 $$ and $$6x+y=3$$ is

A
$$3x-2y=0$$

B
$$2x-3y=0$$

C
$$3x+2y=0$$

D
None of these

Solution

Let us find the Point opposite to origin in the Quadrilateral which will be the intersection of $$x+y=1, 6x+y=3$$

Hence, Solving the Above Equation Simulataneously
$$x=\dfrac{2}{5},y=\dfrac35$$

Now, The Slope of Required Diagonal
$$=\dfrac{\dfrac35-0}{\dfrac25-0}=\dfrac32$$

Hence, Equation of Diagonal
$$(y-0)=\dfrac32(x-0)\Rightarrow 2y=3x\Rightarrow 3x-2y=0$$
Question 3
1 / -0

If the line $$3x-y+1=0$$ and $$x-2y+3=0$$ are equally inclined to the line $$y=mx$$, then the value of $$m$$ is given by

A
$$2m^{2}-7m-7=0$$

B
$$7m^{2}-7m-7=0$$

C
$$7m^{2}-2m-7=0$$

D
$$2m^{2}-7m-2=0$$

Solution
Question 4
1 / -0

The shortest distance of the point $$(h,k)$$ from both the axes are

A
$$|h|,|k|$$

B
$$(h,k)$$

C
$$\sqrt{h^2+k^2},0$$

D
none of these

Solution

Given the point is $$(h,k).$$

Now foot of the perpendicular

from $$(h,k)$$ on x-axis is $$(h,0)$$and

that on y-axis is $$(0,k)$$

Now the shortest distance between

$$ (h,k)$$ and $$(h,0) = \sqrt{(h-h)^{2}+(k-0)^{2}}$$

$$ = |k| $$ = the shortest distance of $$(h,k)$$ from x- axis.

Again the shortest distance between
$$(h,k)$$ and $$(0,k)$$ is

$$ \sqrt{(h-0)^{2}+(k-k)^{2}} = \left | h \right |$$ = The shortest distance of $$(h,k)$$ from y-axis.
Question 5
1 / -0

One vertex of a square $$ABCD$$ is $$A(-1, 1)$$ and the equation of one diagonal $$BD$$ is $$3x+y-8=0$$ then $$C$$=

A
$$(-5, 3)$$

B
$$(5, 3)$$

C
$$(-5, -3)$$

D
$$(2, 5)$$

Solution

Let point $$C (x,y)$$.
Slope of $$AC = \dfrac{{y - 1}}{{x + 1}}$$
and slope of $$BD = 3x + y - 8 = 0$$
$$\begin{array}{l} y=-3x+8\to \left( i \right) \\ \Rightarrow Slope\, \, of\, \, BD\, \, is\, -3 \\ \Rightarrow AC\bot BD \\ \Rightarrow \dfrac { { y-1 } }{ { x+1 } } \times -3=-1 \\ \Rightarrow 3\left( { y-1 } \right) =x+1 \\ \Rightarrow 3y-3=x+1 \\ \Rightarrow 3y-x=4\to \left( { ii } \right) \end{array}$$
Solving equation (i) and (ii)
$$\begin{array}{l} \Rightarrow \dfrac { x }{ { 20 } } =\dfrac { y }{ { 20 } } =\dfrac { 1 }{ { 20 } } \\ \therefore x=2,y=2 \\ \Rightarrow then,\, \, by\, \, { { using } }\, \, mid\, formula,\dfrac { { x-1 } }{ 2 } =2,x=5 \\ \dfrac { { y+1 } }{ 2 } =2,y=3 \end{array}$$
Question 6
1 / -0

If the area of triangle formed by the points $$(2a,b),(a+b,2b+a)$$ and $$(2b,2a)$$ be $$\lambda$$ then the area of the triangle whose vertices are $$(a+b,a-b), (3b-a,b+3a)$$ and $$(3a-b,3b-a)$$ will be

A
$$\dfrac{3}{2} \lambda$$

B
$$3\lambda$$

C
$$4\lambda$$

D
$$none\ of\ these$$

Solution

Area of $$ \triangle \,= \dfrac{1}{2}\left | x_{1}(y_{2}-y_{3})+x_{2}(y_{3}-y_{1})+x_{3}(y_{1}-y_{2}) \right |$$
Given area of triangle whose vertices are $$ (2a,b), (a+b,2b+a), (2b,2a)$$ is $$ \lambda $$
$$\therefore \lambda = \dfrac{1}{2}\left | 2a(2b+a-2a)+(a+b)(2a-b)+2b(b-2b-a) \right | $$
$$ \lambda = \dfrac{1}{2}\left | 4ab+2a^{2}-4a^{2}+2a^{2}-ab+2ab-b^{2}+2b^{2}-4b^{2}-2ab \right |$$
$$ \lambda = \dfrac{1}{2}\left | 3ab - 3b^{2} \right | $$ __ (1)
Now area of triangle whose vertices are $$ (a+b,a-b), (3b-a,b+3a), (3a-b,3b-a)$$ is given by
Area $$ = \dfrac{1}{2}\left | (a+b)(b+3a-3b+a)+(3b-a)(3b-a-a+b)+(3a-b)(a-b-b-3a) \right |$$
$$ = \dfrac{1}{2}\left | (a+b) (-2b+4a)+(3b-a)(4b-2a)+(3a-b)(-2a-2b) \right |$$
$$ = \dfrac{1}{2}\left | -2ab+4a^{2}-2b^{2}+4ab+12b^{2}-6ab-4ab+2a^{2}-6a^{2}-6ab+2ab+2b^{2} \right |$$
$$ = \dfrac{1}{2}\left | 12b^{2}-12ab \right |$$
$$ = \dfrac{4}{2}\left | -(3ab-3b^{2}) \right |$$
$$ = 4.\dfrac{1}{2}\left | 3ab-3b^{2} \right |$$
$$ = 4 \lambda .$$
Question 7
1 / -0

The area of the triangle vertices $$(1,0),(7,0)$$ and $$(4,4)$$ is ___ square units.

A
$$8$$

B
$$10$$

C
$$12$$

D
$$14$$

Solution
Question 8
1 / -0

A triangle with vertices $$(4,0),(-1,-1),(3,5)$$ is:

A
Isosceles and right angled

B
Isosceles but not right angled

C
right angled but not isosceles

D
Neither right angled nor isosceles

Solution

$$\textbf{Step -1: Finding all the side lengths.}$$

$$\text{Let }(4,0)\text{ be point }A,\;(-1,-1)\text{ be point }B,\;(3,5)\text{ be point }C$$

$$AB=\sqrt{(4-(-1))^2+(0-(-1))^2}=\sqrt{5^2+1^2}=\sqrt{26}$$

$$BC=\sqrt{(-1-3)^2+(-1-5)^2}=\sqrt{(-4)^2+(-6)^2}=\sqrt{52}$$

$$CA=\sqrt{(3-4)^2+(5-0)^2}=\sqrt{(-1)^2+(5)^2}=\sqrt{26}$$

$$\therefore AB=CA\text{ Hence, the triangle is isosceles.}$$

$$\textbf{Step -2: Checking whether the triangle is right angled.}$$

$$AB^2=26$$

$$BC^2=52$$

$$CA^2=26$$

$$\Rightarrow AB^2+CA^2=26+26=52=BC^2$$

$$\text{Hence, the triangle is right angled.}$$

$$\textbf{Hence, the triangle with vertices (4,0), (-1,-1), (3,5) is isosceles and right angled.}$$
Question 9
1 / -0

If in a parallelogram $$ABCD$$, the coordinate of $$A, B$$ and $$C$$ are respectively $$(1, 2), (3, 4)$$ and $$(2, 5)$$, then the equation of the diagonal $$BD$$ is:-

A
$$5x + 3y - 11 = 0$$

B
$$3x - 5y + 7 = 0$$

C
$$3x + 5y - 13 = 0$$

D
$$5x - 3y + 9 = 0$$

Solution

Let let coordinates of D are $$X$$ and $$Y$$ then $$X+3=3,Y+4=7$$

$$X=0,Y=3$$

Only above coordinates statisfied by $$5x-3y+9=0$$
Question 10
1 / -0

In the given figure (not drawn to scale), the coordinates of P, Q and R are (3, 0) (0, 5) and (0, -5) respectively. Find the area of $$\Delta PQR$$.

A
20 sq. units

B
40 sq. units

C
15 sq. units

D
25/2 sq. units

Solution