Q: What is the angle between two lines whose slopes are $m1 = 2$ and $m2 = 3$?

The acute angle $\theta$ between two lines with slopes $m1$ and $m2$ is given by $\tan \theta = \left| \frac{m2 - m1}{1 + m1 m2} \right|$. Substituting $m1 = 2$ and $m2 = 3$: $\tan \theta = \left| \frac{3 - 2}{1 + (2)(3)} \right| = \left| \frac{1}{7} \right| = \frac{1}{7}$. Thus, $\theta = \tan^{-1}

Question 1

What is the slope of a line passing through the points $(2, 3)$ and $(4, 7)$?

Accepted Answer

The slope $m$ of a line passing through two points $(x1, y1)$ and $(x2, y2)$ is given by the formula $m = \frac{y2 - y1}{x2 - x1}$. Substituting the given points $(2, 3)$ and $(4, 7)$: $m = \frac{7 - 3}{4 - 2} = \frac{4}{2} = 2$. Therefore, the correct option is B.

Question 2

What is the angle between two lines whose slopes are $m1 = 2$ and $m2 = 3$?

Accepted Answer

The acute angle $	heta$ between two lines with slopes $m1$ and $m2$ is given by $	an 	heta = \left| \frac{m2 - m1}{1 + m1 m2} ight|$. Substituting $m1 = 2$ and $m2 = 3$: $	an 	heta = \left| \frac{3 - 2}{1 + (2)(3)} ight| = \left| \frac{1}{7} ight| = \frac{1}{7}$. Thus, $	heta = 	an^{-1}

Question 3

If three points $(h, 0)$, $(a, b)$, and $(0, k)$ are collinear, then which of the following relations is true?

Accepted Answer

If three points are collinear, the slope of the line segment joining the first two points is equal to the slope of the line segment joining the last two points. Slope between $(h, 0)$ and $(a, b)$ is $\frac{b - 0}{a - h}$. Slope between $(a, b)$ and $(0, k)$ is $\frac{k - b}{0 - a}$. Equating them:

Question 4

What is the equation of the line passing through $(-4, 3)$ with a slope of $1/2$?

Accepted Answer

Using the point-slope form: $y - y1 = m(x - x1)$. Here, $x1 = -4, y1 = 3, m = 1/2$. Substituting these values: $y - 3 = \frac{1}{2}(x - (-4)) \implies y - 3 = \frac{1}{2}(x + 4) \implies 2(y - 3) = x + 4 \implies 2y - 6 = x + 4 \implies x - 2y + 10 = 0$. Therefore, the correct option is B.

Question 5

What is the equation of the line that makes intercepts $-3$ and $2$ on the x-axis and y-axis respectively?

Accepted Answer

The intercept form of a line is $\frac{x}{a} + \frac{y}{b} = 1$, where $a$ is the x-intercept and $b$ is the y-intercept. Given $a = -3$ and $b = 2$. Substituting these: $\frac{x}{-3} + \frac{y}{2} = 1 \implies \frac{-2x + 3y}{6} = 1 \implies -2x + 3y = 6 \implies 2x - 3y + 6 = 0$. Therefore, the co

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