Question

14. A (2, 3) and B(-2, 5) are two given points.Find (i) the gradient of AB:
(¡¡)The equation of AB:
(¡¡¡) The co-ordinates of the points, where AB intersects x-axis

Answer 1

Answer:

According to question A(2, 3)and B (-2, 5) are the given points.

I) we have to find the gradient / slope of AB:-)

As if

coordinates of two points are( x1 , y1 )and (x2 , y2)

slope of the line joining these points will be

$\dfrac{y _2 - y _1 }{x _2 - x_1 }$

So,

slope will be

$\frac{ 5 - 3}{ - 2 - 2} \\ \\ = \frac{2}{ - 4} \\ \\ = - \frac{ 1}{2}$

ii)

We have slope point form find eq. of any line using its slope and one passing point.

i.e.

$y - y_1 = m(x - x_1)$

where m is the slope of the line.

So,

Equation of this line will be,

$(y - 3) = - \dfrac{1}{2} (x - 2) \\ \\ 2y - 6 = - x + 2 \\ \\ x + 2y = 8$

III)

where the line cuts x-axis then Y co-ordinate will be zero

so,

$x = 8$

So, the point where this line cut x-axis will be (8,0)

Answer 2

Step-by-step explanation:

Answer:

According to question A(2, 3)and B (-2, 5) are the given points.

I) we have to find the gradient / slope of AB:-)

As if

coordinates of two points are( x1 , y1 )and (x2 , y2)

slope of the line joining these points will be

\dfrac{y _2 - y _1 }{x _2 - x_1 }x2−x1y2−y1

So,

slope will be

\begin{gathered} \frac{ 5 - 3}{ - 2 - 2} \\ \\ = \frac{2}{ - 4} \\ \\ = - \frac{ 1}{2} \end{gathered}−2−25−3=−42=−21

ii)

We have slope point form find eq. of any line using its slope and one passing point.

i.e.

y - y_1 = m(x - x_1)y−y1=m(x−x1)

where m is the slope of the line.

So,

Equation of this line will be,

\begin{gathered}(y - 3) = - \dfrac{1}{2} (x - 2) \\ \\ 2y - 6 = - x + 2 \\ \\ x + 2y = 8\end{gathered}(y−3)=−21(x−2)2y−6=−x+2x+2y=8

III)

where the line cuts x-axis then Y co-ordinate will be zero

so,

x = 8x=8

So, the point where this line cut x-axis will be (8,0)