Answer :

Let $A B C D$ be the given quadrilateral with vertices $ A($ -4, 5 $), B(0,7), C ( 5, - $5 $)$, and $D$ (4, 2).

Then, by plotting $A, B, C$, and $D$ on the Cartesian plane and joining $A B, B C, C D$, and $D A$, the given quadrilateral can be drawn as

To find the area of quadrilateral $A B C D$, we draw one diagonal, say $A C$.

Accordingly, area $(A B C D)=area(\triangle A B C)+area(\triangle A C D)$

We know that the area of a triangle whose vertices are $(x_1, y_1),(x_2, y_2)$, and $(x_3, y_3)$ is

$\dfrac{1}{2}|x_1(y_2-y_3)+x_2(y_3-y_1)+x_3(y_1-y_2)|$

Therefore, area of $\triangle ABC$ $=\dfrac{1}{2}|-4(7+5)+0(-5-5)+5(5-7)|$ unit $^{2}$

$=\dfrac{1}{2}|-4(12)+5(-2)|$ unit $^{2}$

$=\dfrac{1}{2}|-48-10|$ unit $^{2}$

$=\dfrac{1}{2}|-58|$ unit $^{2}$

$=\dfrac{1}{2} \times 58$ unit $^{2}$

$=29$ unit $^{2}$

Area of $\triangle A C D$

$=\dfrac{1}{2}|-4(-5+2)+5(-2-5)+(-4)(5+5)|$ unit $^{2}$

$=\dfrac{1}{2}|-4(-3)+5(-7)-4(10)|$ unit $^{2}$

$=\dfrac{1}{2}|12-35-40|$ unit $^{2}$

$=\dfrac{1}{2}|-63|$ unit $^{2}$

$=\dfrac{63}{2}$ unit $^{2}$

Thus, area (ABCD)

$ =(29+\dfrac{63}{2}) \text{ unit }^{2}=\dfrac{58+63}{2} \text{ unit }^{2}=\dfrac{121}{2} \text{ unit }^{2} $

Answer :

Let $A B C$ be the given equilateral triangle with side $2 a$.

Accordingly, $AB=BC=CA=2 a$

Assume that base $BC$ lies along the $y$-axis such that the mid-point of $BC$ is at the origin.

i.e., $BO=OC=a$, where $O$ is the origin.

Now, it is clear that the coordinates of point $C$ are $(0, a)$, while the coordinates of point $B$ are $(0, - a)$.

It is known that the line joining a vertex of an equilateral triangle with the mid-point of its opposite side is perpendicular.

Hence, vertex A lies on the $y$-axis.

On applying Pythagoras theorem to $\triangle A O C$, we obtain

$ \begin{aligned} & (AC)^{2}=(OA)^{2}+(OC)^{2} \\ & \Rightarrow(2 a)^{2}=(OA)^{2}+a^{2} \\ & \Rightarrow 4 a^{2} - a^{2}=(OA)^{2} \\ & \Rightarrow(OA)^{2}=3 a^{2} \\ & \Rightarrow OA=\sqrt{3} a \end{aligned} $

$\therefore$ Coordinates of point $A=( \pm \sqrt{3} a, 0)$

Thus, the vertices of the given equilateral triangle are $(0, a),(0, -{ } a)$, and $(\sqrt{3} a, 0)$ or $(0, a),(0, - a)$, and $(-\sqrt{3} a, 0)$.

Answer :

The given points are $P(x_1, y_1)$ and $Q(x_2, y_2)$.

(i) When PQ is parallel to the $y$-axis, $x_1=x_2$.

In this case, distance between $P$ and $Q=\sqrt{(x_2-x_1)^{2}+(y_2-y_1)^{2}}$

$ \begin{aligned} & =\sqrt{(y_2-y_1)^{2}} \\ & =|y_2-y_1| \end{aligned} $

(ii) When PQ is parallel to the $x$-axis, $y_1=y_2$.

In this case, distance between $P$ and $Q=\sqrt{(x_2-x_1)^{2}+(y_2-y_1)^{2}}$

$ \begin{aligned} & =\sqrt{(x_2-x_1)^{2}} \\ & =|x_2-x_1| \end{aligned} $

Answer :

Let $(a, 0)$ be the point on the $x$ axis that is equidistant from the points $(7,6)$ and $(3,4)$.

Accordingly, $\sqrt{(7-a)^{2}+(6-0)^{2}}=\sqrt{(3-a)^{2}+(4-0)^{2}}$

$\Rightarrow \sqrt{49+a^{2}-14 a+36}=\sqrt{9+a^{2}-6 a+16}$

$\Rightarrow \sqrt{a^{2}-14 a+85}=\sqrt{a^{2}-6 a+25}$

On squaring both sides, we obtain

$a^{2}$ - $14 a+85=a^{2}- 6 a+25$

$\Rightarrow a - 14 a+6 a=25 - 85$

$\Rightarrow $ - $8 a=- 60$

$\Rightarrow a=\dfrac{60}{8}=\dfrac{15}{2}$

Thus, the required point on the $x$-axis is

$ (\dfrac{15}{2}, 0) $

Answer :

The coordinates of the mid-point of the line segment joining the points

$P(0, - 4)$ and $B(8,0)$ are

$ (\dfrac{0+8}{2}, \dfrac{-4+0}{2})=(4,-2) $

It is known that the slope $(m)$ of a non-vertical line passing through the points $(x_1, y_1)$ and $(x_2, y_2)$ is given by $m=\dfrac{y_2-y_1}{x_2-x_1}, x_2 \neq x_1$

Therefore, the slope of the line passing through $(0,0)$ and $ ( 4,-2) $ is

$\dfrac{-2-0}{4-0}=\dfrac{-2}{4}=-\dfrac{1}{2}$.

Hence, the required slope of the line is $-\dfrac{1}{2}$.

Answer :

The vertices of the given triangle are A $(4,4), B(3,5)$, and C (-1, -1).

It is known that the slope $(m)$ of a non-vertical line passing through the points $(x_1, y_1)$ and $(x_2, y_2)$ is given by $m=\dfrac{y_2-y_1}{x_2-x_1}, x_2 \neq x_1$

$\therefore$ Slope of $AB(m_1)=\dfrac{5-4}{3-4}=-1$

Slope of $BC(m_2)=\dfrac{-1-5}{-1-3}=\dfrac{-6}{-4}=\dfrac{3}{2}$

Slope of CA $(m_3)=\dfrac{4+1}{4+1}=\dfrac{5}{5}=1$

It is observed that $m_1 m_3= - 1 $

This shows that line segments $A B$ and $C A$ are perpendicular to each other i.e., the given triangle is right-angled at $A(4,4)$.

Thus, the points $(4,4),(3,5)$, and (1, 1 ) are the vertices of a right-angled triangle.

Answer :

If a line makes an angle of $30^{\circ}$ with the positive direction of the $y$-axis measured anticlockwise, then the angle made by the line with the positive direction of the $x$-axis measured anticlockwise is $90^{\circ}+30^{\circ}=120^{\circ}$.

Thus, the slope of the given line is $\tan 120^{\circ}=\tan (180^{\circ} - 60^{\circ})=- \tan 60^{\circ}=-\sqrt{3}$

Answer :

Let points $( - 2, -1), (4,0),(3,3)$, and $( - 3,2 )$ be respectively denoted by A, B, C, and $D$.

Slope of $A B=\dfrac{0+1}{4+2}=\dfrac{1}{6}$

Slope of $CD=\dfrac{2-3}{-3-3}=\dfrac{-1}{-6}=\dfrac{1}{6}$

$\Rightarrow$ Slope of $A B=$ Slope of $C D$

$\Rightarrow A B$ and $C D$ are parallel to each other.

Now, slope of $BC=\dfrac{3-0}{3-4}=\dfrac{3}{-1}=-3$

Slope of $AD=\dfrac{2+1}{-3+2}=\dfrac{3}{-1}=-3$

$\Rightarrow$ Slope of $BC=$ Slope of $AD$

$\Rightarrow B C$ and $A D$ are parallel to each other.

Therefore, both pairs of opposite sides of quadrilateral $A B C D$ are parallel. Hence, $A B C D$ is a parallelogram.

Thus, points (-2, -1), $(4,0),(3,3)$, and (-3,2) are the vertices of a parallelogram.

Answer :

The slope of the line joining the points $(3,-1)$ and $(4,-2)$. is

$ m=\dfrac{-2-(-1)}{4-3}=-2+1=-1 $

Now, the inclination $(\theta)$ of the line joining the points $(3, -1 )$ and $(4 ,-2)$ is given by $\tan \theta=- 1$

$\Rightarrow \theta=(90^{\circ}+45^{\circ})=135^{\circ}$

Thus, the angle between the $x$-axis and the line joining the points $(3,- 1)$ and $(4, -2 )$ is $ 135^{\circ} $.

Answer :

Let $m_1$ and $m$ be the slopes of the two given lines such that $m_1=2 m$.

We know that if $\theta$ isthe angle between the lines $l_1$ and $I_2$ with slopes $m_1$ and $m_2$, then

$ \tan \theta=|\dfrac{m_2-m_1}{1+m_1 m_2}| $

It is given that the tangent of the angle between the two lines is $\dfrac{1}{3}$.

$\therefore \dfrac{1}{3}=|\dfrac{m-2 m}{1+(2 m) \cdot m}|$

$\Rightarrow \dfrac{1}{3}=|\dfrac{-m}{1+2 m^{2}}|$

$\Rightarrow \dfrac{1}{3}=\dfrac{-m}{1+2 m^{2}}$ or $\dfrac{1}{3}=-(\dfrac{-m}{1+2 m^{2}})=\dfrac{m}{1+2 m^{2}}$

If $m=$ - 1 , then the slopes of the lines are - 1 and - 2 .

If $m=-\dfrac{1}{2}$, then the slopes of the lines are $-\dfrac{1}{2}$ and -1.

Case II

$\dfrac{1}{3}=\dfrac{m}{1+2 m^{2}}$

$\Rightarrow 2 m^{2}+1=3 m$

$\Rightarrow 2 m^{2}-3 m+1=0$

$\Rightarrow 2 m^{2}-2 m-m+1=0$

$\Rightarrow 2 m(m-1)-1(m-1)=0$

$\Rightarrow(m-1)(2 m-1)=0$

$\Rightarrow m=1$ or $m=\dfrac{1}{2}$

If $m=1$, then the slopes of the lines are 1 and 2 .

If $m=\dfrac{1}{2}$, then the slopes of the lines are $\dfrac{1}{2}$ and 1

Hence, the slopes of the lines are - 1 and - 2 or $\quad-\dfrac{1}{2}$ and - 1 or 1 and 2 or $\dfrac{1}{2}$ and 1 .

Answer :

The slope of the line passing through $(x_1, y_1)$ and $(h, k)$ is $\dfrac{k-y_1}{h-x_1}$.

It is given that the slope of the line is $m$.

$\therefore \dfrac{k-y_1}{h-x_1}=m$

$\Rightarrow k-y_1=m(h-x_1)$

Hence, $k-y_1=m(h-x_1)$