Slope and y-intercept

A line parallel to the \(y\)-axis is vertical, so its equation has the form \(x=\text{constant}\). Since the point has \(x\)-coordinate 3, the line is \(x=3\). \(y=-2\) is horizontal. Exam tip: a vertical line has undefined slope.

A line in the form y = mx + c has y-intercept c. Here, the y-intercept is u + 6. So, u + 6 = 16, giving u = 10. Therefore, the slope is 3u - 2 = 3(10) - 2 = 28. Option 26 can result from an incorrect subtraction. Exam tip: to find the y-intercept, put x = 0.

The direct answer is option A: y-intercept 11 and slope \(-\frac{5}{2}\). First isolate y: \(10x+4y-44=0\). Move the x-term and constant: \(4y=-10x+44\). Divide every term by 4: \(y=-\frac{10}{4}x+\frac{44}{4}=-\frac{5}{2}x+11\). Comparing with \(y=mx+c\), the slope is \(m=-\frac{5}{2}\) and the y-intercept is \(c=11\). Option A is correct. Option B has both signs reversed. Option C treats original coefficients as the answer and misses the required rearrangement. Option D incorrectly swaps the coefficients and does not give slope-intercept form. A y-intercept is the value of y when x is zero; here it is 11. Memory cue: rewrite as \(y=mx+c\); the coefficient of x is slope and the constant is the y-intercept.

In the form y=mx+c, the y-intercept is c because substituting x=0 gives y=c. The first line has y-intercept b, while the second has y-intercept 13. Since they meet the y-axis at the same point, b=13. The values 8 and −6 are slopes, not y-intercepts. Exam tip: put x=0 to find a y-intercept.

In the form y = mx + c, the coefficient of x, m, is the slope. Since the two lines have equal slopes, a = 10. Substituting x = 3 in the first line gives y = 10(3) - 12 = 18. The value 20 would not account correctly for the constant term -12. Exam tip: for lines with equal slopes, compare the coefficients of x.

The y-intercept is the value of y when x=0, so c=7. Substituting x=-3 gives y=-11(-3)+7=33+7=40. The value 44 can result from incorrectly adding 11 to 33. Exam tip: identify the constant term from the y-intercept and carefully handle signs when multiplying negative numbers.

In \(y=mx+c\), \(m\) is the slope and \(c\) is the y-intercept. For \(y=2x-5\), \(m=2>0\) and \(c=-5<0\). Although \(y=-2x-5\) has a negative intercept, its slope is negative too. Exam tip: check the coefficient of x first.

In \(y=mx+c\), \(m=-3\) is negative, so the line falls from left to right. Also, \(c=5\) is the y-intercept. Exam tip: identify the sign of slope and the constant term separately.

In slope-intercept form \(y=mx+c\), \(m\) is the slope and \(c\) is the y-intercept. For \(y=2x-3\), \(m=2\) and \(c=-3\). Option B has a negative slope. Exam tip: identify the coefficient of \(x\) and the constant term first.

In the equation of a line, the slope is represented by \(m\). Given \(m+c=21\) and \(c=8\), we get \(m+8=21\). Hence, \(m=21-8=13\). Therefore, the correct answer is 13. Note that 8 is \(c\), the y-intercept, not the slope. Exam tip: in \(y=mx+c\), the coefficient \(m\) is always the slope.

In the form \(y=mx+c\), \(m\) is the slope and \(c\) is the \(y\)-intercept. Given slope \(m=-7\), substitute it into \(m+c=5\): \(-7+c=5\). Hence, \(c=12\). Choosing \(-12\) would result from a sign error. Exam tip: add \(7\) to both sides to isolate \(c\).

For the given point, substitute \(x=12\) and \(y=19\): \(19=\frac{5}{6}\times 12+c\). Since \(\frac{5}{6}\times 12=10\), we get \(19=10+c\), so \(c=9\). Option 10 is the value of the slope term, not the y-intercept. Exam tip: in \(y=mx+c\), substitute both coordinates of the given point to find \(c\).

The given line \(y=-3x+5\) has slope \(-3\). Parallel lines must have the same slope. Also, in \(y=mx+c\), \(c\) is the y-intercept, which must be \(-2\) here. Therefore, the required equation is \(y=-3x-2\). Option C has the correct slope but its y-intercept is \(+2\). Exam tip: verify the slope first and then check the constant term for the y-intercept.

In the form \(y=mx+c\), the y-intercept is the constant term \(c\). Here, \(a+4=13\), so \(a=9\). Therefore, the slope is \(m=6a-5=6(9)-5=49\). Choosing 50 would result from incorrectly handling the subtraction of 5. Exam tip: first find the parameter using the y-intercept, then substitute it into the coefficient of \(x\).

In the form \(y=mx+c\), the coefficient of \(x\) is the slope. Thus, \(b-13=4\), giving \(b=17\). The \(y\)-intercept is the constant term \(3b+1\): \(3(17)+1=52\). Therefore, 52 is correct. The value 51 would result from using only \(3b\) and missing the \(+1\). Exam tip: first find the parameter from the slope, then substitute it into the constant term.

In the form \(y=mx+c\), \(m\) is the slope and \(c\) is the y-intercept. Equal slopes give \(t+5=14\), so \(t=9\). The y-intercept of the second line is \(2t-7\); hence \(2(9)-7=11\). Option 9 is the value of \(t\), not the y-intercept. Exam tip: first equate the slope coefficients to find the parameter, then substitute it into the constant term.

In the form \(y=mx+c\), \(c\) is the \(y\)-intercept. Since the two lines have the same \(y\)-intercept, \(v-9=8\). Thus, \(v=17\). The slope of the second line \(y=(v+4)x+8\) is \(v+4=17+4=21\). The value 19 would be \(v+2\), not the required slope. Exam tip: in \(y=mx+c\), the coefficient of \(x\) is the slope.

The slope is \(-\frac{7}{4}\). Thus, when \(x\) increases by 4 units, \(y\) decreases by 7 units. For a decrease of 42 in \(y\), \(\frac{7}{4}\Delta x=42\). Hence \(\Delta x=42\times\frac{4}{7}=24\). If 28 were chosen, the decrease in \(y\) would be \(\frac{7}{4}\times 28=49\), not 42. Exam tip: In change-based questions, use the slope; the constant term 18 does not affect the change.

At \(d=0\), \(T=80\), so ₹80 is the y-intercept or fixed initial fee. Increasing \(d\) by 1 km adds ₹12.5, so it is the slope. Treating ₹12.5 as the intercept confuses the coefficient with the constant. Exam tip: set the input to zero to find the intercept.

Given \(c=-15\) and \(m-c=26\). Substituting \(c\) gives \(m-(-15)=26\), so \(m+15=26\). Hence, \(m=11\), and the slope is 11. For example, 10 is not correct because \(10-(-15)=25\), not 26. Exam tip: subtracting a negative number is the same as adding it.

In the line form \(y=mx+c\), \(c\) is the value of the \(y\)-intercept. Substituting \(m=-5\) into \(3c+m=31\) gives \(3c-5=31\). Hence, \(3c=36\), so \(c=12\). Therefore, the \(y\)-intercept is 12. Choosing 11 would be incorrect because it does not satisfy \(3c-5=31\). Exam tip: substitute the given value of \(m\) into the relation first, then solve for \(c\).

\(x=3\) is a vertical line. Here \(\Delta x=0\), so \(m=\frac{\Delta y}{\Delta x}\) is undefined. It does not meet the \(y\)-axis, where \(x=0\). In contrast, \(y=3\) has zero slope. Exam tip: a line of the form \(x=\text{constant}\) is vertical.

Dividing both sides of the equation by 8 gives y=3x-8. For the y-intercept, put x=0, which gives y=-8. Therefore, the y-intercept is -8. Although -64 is the constant term in the original equation, it must be divided by 8 before identifying the y-intercept. Exam tip: In the form y=mx+c, c is the y-intercept.

In \(y=mx+c\), \(m\) is the slope and \(c\) is the y-intercept. For option A, \(m=3\) is positive and \(c=-5\) is negative. Option B has a negative intercept but its slope is negative. Exam tip: check the signs of \(m\) and \(c\) separately.

Substitute the point \((-5,12)\) into \(y=mx-13\): \(12=m(-5)-13\). Hence, \(25=-5m\), so \(m=-5\). Therefore, the correct option is \(\mathbf{-5}\). If \(m=5\), the equation gives \(y=-38\) when \(x=-5\), so it cannot pass through the given point. Exam tip: To find an unknown slope parameter, substitute the coordinates of the given point into the line equation.