Answer:
1)a. j and k
2) d. \(y = \frac{-1}[3}x+3\)
3)a. (6, 12)
4)d. 10
Step-by-step explanation:
1) Lines are said to parallel if their slopes are same.
General form of line = \(y = mx+c\) --1
where m is the slope
On comparing all lines with 1
So, Slope of line j = \(\frac{1}{4}\)
Slope of line k = \(\frac{1}{4}\)
Slope of line l = 3
Slope of line m = 4
Slope of line j and k are same
So, Option a is correct.
a. j and k
2) Find the equation of a line perpendicular to y-3x=-8 that passes through the point (3, 2).
y=-8+3x
On comparing with 1
Slope of given line is 3
Now slope of a line which is perpendicular to the given line
Two lines are said to be perpendicular if the product of their slopes is -1
So, \(3 \times \frac{-1}{3}=-1\)
So, slope of perpendicular line is \(\frac{-1}{3}\)
General form of line = \(y = mx+c\) -1
Substitute m = \(\frac{-1}{3}\) and passing points (3,2)
\(2 = 3\times \frac{-1}[3}+c\)
\(3=c\)
So, Now substitute value of m and c in 1
\(y = \frac{-1}[3}x+3\)
Hence the equation of a line perpendicular to y-3x=-8 that passes through the point (3, 2) is \(y = \frac{-1}[3}x+3\)
Option d is correct.
3) Find the point, M, that divide segment segment AB into a ratio of 5:2 if A is at (1, 2) and B is at (8, 16).
m:n=5:2
\((x_1,y_1)=(1,2)\)
\((x_2,y_2)=(8,16)\)
To find coordinates of M we will sue section formula:
\(x=\frac{mx_2+nx_1}{m+n}\) and \(y=\frac{my_2+ny_1}{m+n}\)
\(x=\frac{5(8)+2(1)}{5+2}\) and \(y=\frac{5(16)+2(2)}{5+2}\)
\(x=\6\) and \(y=12\)
Thus the coordinates of M is (6,12)
Hence Option A is correct.
4). Find the distance between (4, 2) and (-4, -4).
\(d=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}\)
\((x_1,y_1)=(4,2)\)
\((x_2,y_2)=(-4,-4)\)
Substitute the values in the formula :
\(d=\sqrt{(-4-4)^2+(-4-2)^2}\)
\(d=\sqrt{(-8)^2+(-6)^2}\)
\(d=\sqrt{64+36}\)
\(d=\sqrt{100}\)
\(d=10\)
Thus the distance is 10
Hence Option D is correct.
