(A) Given the curve $y = x^2 + 2x$.For $x_1 = 1$,$y_1 = (1)^2 + 2(1) = 1 + 2 = 3$. So,the first point is $(1, 3)$.For $x_2 = 3$,$y_2 = (3)^2 + 2(3) =

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(A) Given the curve $y = x^2 + 2x$.For $x_1 = 1$,$y_1 = (1)^2 + 2(1) = 1 + 2 = 3$. So,the first point is $(1, 3)$.For $x_2 = 3$,$y_2 = (3)^2 + 2(3) = 9 + 6 = 15$. So,the second point is $(3, 15)$.The gradient $m$ of the line joining $(x_1, y_1)$ and $(x_2, y_2)$ is given by $m = \frac{y_2 - y_1}{x_2 - x_1}$.Substituting the values,$m = \frac{15 - 3}{3 - 1} = \frac{12}{2} = 6$.

Class 11 Mathematics Straight Lines Slope of line, Equation of line in different forms

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The gradient of the line joining the points on the curve $y = x^2 + 2x$ whose abscissae are $1$ and $3$,is

A
$6$
B
$5$
C
$4$
D
$3$

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Straight Lines

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Slope of line, Equation of line in different forms

If the line $6x - 7y + 8 + \lambda(3x - y + 5) = 0$ is parallel to the $y$-axis,then $\lambda$ is equal to

MediumKCET 2013

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The equation of the line passing through $(1, 2)$ and parallel to the line $y = 3x - 1$ is

Easy

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Two points $(a, 0)$ and $(0, b)$ are joined by a straight line. Another point on this line is:

Easy

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$A$ line meets the $x$-axis and $y$-axis at the points $A$ and $B$ respectively. If the midpoint of $AB$ is $(x_1, y_1)$,then the equation of the line is:

Medium

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Reduce the following equation into intercept form and find its intercepts on the axes: $4x - 3y = 6$.

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The gradient of the line joining the points on the curve $y = x^2 + 2x$ whose abscissae are $1$ and $3$,is

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If the line $6x - 7y + 8 + \lambda(3x - y + 5) = 0$ is parallel to the $y$-axis,then $\lambda$ is equal to

The equation of the line passing through $(1, 2)$ and parallel to the line $y = 3x - 1$ is

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$A$ line meets the $x$-axis and $y$-axis at the points $A$ and $B$ respectively. If the midpoint of $AB$ is $(x_1, y_1)$,then the equation of the line is:

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