Question

plzz solve 3rd and 4th.
don't post irrelevant answer. ​

Answer 1

3)

Given:

• A(5,2), B(-7,2) and C(6,3)

Find:

• Check Whether these points are colinear points or not.

Solution:

we, know that Collinear Points are those points that lies on the same line.

So,

$\setlength{\unitlength}{1cm} \begin{picture}(6,6) \linethickness{0.5mm}\put(2,1){\line(1,0){4}}\multiput(2,0.8)(2,0) {3}{\line(0,1){0.4}} \put(3.5,1.3){\bf B( - 7,2)} \put(1.5,1.3){\bf A(5,2)} \put(5.7,1.3){\bf C(6,3)} \end{picture}$

Here, we have to prove

AC = AB + BC

So,

we, know that

$\boxed{ \sf Distance \: Formula = \sqrt{ {(x_{2} - x_{1})}^{2} + {(y_{2} - y_{1})}^{2}} \: }$

So,

$: \to \sf AC = \sqrt{ {(x_{2} - x_{1})}^{2} + {(y_{2} - y_{1})}^{2}} \:$

where,

• $\sf x_1 = 5$
• $\sf x_2 = 6$
• $\sf y_1 = 2$
• $\sf y_2 = 3$

So,

$: \to \sf AC = \sqrt{ {(6 - 5)}^{2} + {(3 - 2)}^{2}} \:$

$: \to \sf AC = \sqrt{ {(1)}^{2} + {(1)}^{2}} \:$

$: \to \sf AC = \sqrt{1+ 1} \:$

$: \to \sf AC = \sqrt{2} \:$

$: \to \sf AC = \sqrt{2} units \:$

Now,

$: \to \sf AB = \sqrt{ {(x_{2} - x_{1})}^{2} + {(y_{2} - y_{1})}^{2}} \:$

where,

• $\sf x_1 = 5$
• $\sf x_2 = -7$
• $\sf y_1 = 2$
• $\sf y_2 = 2$

So,

$: \to \sf AB = \sqrt{ {( - 7- 5)}^{2} + {(2 - 2)}^{2}} \:$

$: \to \sf AB = \sqrt{ {( -12)}^{2} + {(0)}^{2}} \:$

$: \to \sf AB = \sqrt{144 + 0} \:$

$: \to \sf AB = \sqrt{144} \:$

$: \to \sf AB = 12units \:$

Now,

$: \to \sf BC = \sqrt{ {(x_{2} - x_{1})}^{2} + {(y_{2} - y_{1})}^{2}} \:$

where,

• $\sf x_1 = -7$
• $\sf x_2 = 6$
• $\sf y_1 = 2$
• $\sf y_2 = 3$

So,

$: \to \sf BC = \sqrt{ {(6 - ( - 7))}^{2} + {(3 - 2)}^{2}} \:$

$: \to \sf BC = \sqrt{ {(6 + 7)}^{2} + {(3 - 2)}^{2}} \:$

$: \to \sf BC = \sqrt{ {(13)}^{2} + {(1)}^{2}} \:$

$: \to \sf BC = \sqrt{169 + 1} \:$

$: \to \sf BC = \sqrt{170} units\:$

Now,

we, have to show

$\sf \dashrightarrow AC = AB + BC$

where,

• AC = √2 units
• AB = 12 units
• BC = √(170) units

So,

$\sf \dashrightarrow \sqrt{2} \not = 12 + \sqrt{170}$

Here,

AC ≠ AB + BC

Hence, the points are non-collinear

______________________________

4)

Given:

• Point (-4,9)

Find:

• Perpendicular Distance of the point from the y-axis

Solution:

we, know

If a point is defined as (x,y), then the distance between y-axis and the point (x,y) is |x| and the distance between x-axis and the point (x,y) is |y|

The given point is (-4,9)

This point is -4 units left from y-axis and the 9 units below the x-axis

It means that the distance between y-axis and the point is -4 and distance between x-axis and the point is 9

Therefore, Perpendicular Distance of the point (-4,9) from the y-axis is -4 units