Find the attached solution

1. The equation of a straight line is given by 𝑦 = 𝑚𝑥 + 𝑐, where 𝑚 is the slope of the line
and 𝑐 is the intercept.
It is given that the slope of the line is 6, so 𝑚 = 6.
So, the equation of the line become, 𝑦 = 6𝑥 + 𝑐.
This line is passing through the point (2,16)
This information will give the value of other parameter 𝑐 as:
Putting the value of 𝑥 = 2 and 𝑦 = 16 in the equation of line,
16 = 6 × 2 + 𝑐
Simplifying the above equation,
16 = 12 + 𝑐 ⇒ 𝑐 = 16 − 12 = 4
Finally, the equation of the line becomes
𝑦 = 6𝑥 + 4
This is the required line that has a slope 6 and passes through the point (2,16)
2. The equation of a straight line is given by 𝑦 = 𝑚𝑥 + 𝑐, where 𝑚 is the slope of the line
and 𝑐 is the intercept.
It is given that the line is perpendicular to another line 3𝑦 − 12𝑥 = 22. Rearranging the
line in standard form,
3𝑦 − 12𝑥 = 22.
⇒ 3𝑦 = 12𝑥 + 22
12
22
⇒𝑦=
𝑥+
3
3
22
⇒ 𝑦 = 4𝑥 +
3
Thus, slope of this line is 𝑚1 = 4
If two lines are perpendicular to each than product of there slopes are equal to -1
𝑚𝑚1 = −1
Since, the slope of other line is obtained as 𝑚1 = 4, the slope of first line will be
𝑚 × 4 = −1
−1
⇒𝑚=
4
−1
So, the equation of the line become, 𝑦 = 4 𝑥 + 𝑐.
This line is passing through the point (4,25)
This information will give the value of other parameter 𝑐 as:
Putting the value of 𝑥 = 4 and 𝑦 = 25 in the equation of line,
−1
24 =
×4+𝑐
4
Simplifying the above equation,
24 = −1 + 𝑐 ⇒ 𝑐 = 24 + 1 = 25
Finally, the equation of the li...