All lines, except those parallel to the \(x\)-axis or the \(y\)-axis, meet both coordinate axes. Suppose that a line \(l\) passes through \((a,0)\) and \((0,b)\). Then \(a\) is the \(x\)-intercept and \(b\) is the \(y\)-intercept of \(l\). The intercepts \(a\) and \(b\) can be positive, negative or zero. All lines through the origin have \(a=0\) and \(b=0\). Equations of linesOne of the axioms of Euclidean geometry is that two points determine a line. In other words, there is a unique line through any two fixed points. This idea translates to coordinate geometry and, as we shall see, all points on the line through two points satisfy an equation of the form \(ax+by+c=0\), with \(a\) and \(b\) not both 0. Conversely, any `linear equation' \(ax+by+c=0\) is the equation of a (straight) line. This is called the general form of the equation of a line. Point–gradient formConsider the line \(l\) which passes through the point \((x_1,y_1)\) and has gradient \(m\). Detailed description Let \(P(x,y)\) be any point on \(l\), except for \((x_1,y_1)\). Then \[ m = \dfrac{y-y_1}{x-x_1}, \] and so \[ y-y_1 = m(x-x_1). \]This equation is called the point–gradient form of the equation of the line \(l\). Suppose that \((x_1,y_1)=(0,c)\). Then the equation is \(y-c=mx\) or, equivalently, \(y=mx+c\). This is often called the gradient–intercept form of the equation of the line. Line through two pointsTo find the equation of the line through two given points \((x_1,y_1)\) and \((x_2,y_2)\), first find the gradient \[ m=\dfrac{y_2-y_1}{x_2-x_1} \] and then use the point–gradient form \(y-y_1 = m(x-x_1)\).A special case is the line through \((a,0)\) and \((0,b)\), where \(a,b \ne 0\). In this case, the gradient is \[ m = \dfrac{b-0}{0-a} = -\dfrac{b}{a}. \]Thus the equation of the line is SummaryThere are four different forms of the equation of a straight line which we have considered.
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The equation of the vertical line through \((3,4)\) is \(x=3\). The equation of the horizontal line through \((3,4)\) is \(y=4\). Detailed description Next page - Content - Parallel, intersecting and perpendicular lines
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Concept: Line: Line in slope form is given as, \(\frac{{\rm{x}}}{{\rm{a}}} + \frac{{\rm{y}}}{{\rm{b}}} = 1\) where a is x-intercept and b is y-intercept. Calculation: Let the equation of line be \(\frac{{\rm{x}}}{{\rm{a}}} + \frac{{\rm{y}}}{{\rm{b}}} = 1\) ----(1) Where a is x-intercept and b is y-intercept. The line is passing through (2,3). So, (2,3) should satisfy the equation of line. Put x = 2 & y = 3 in equation (1) \(\frac{2}{{\rm{a}}} + \frac{3}{{\rm{b}}} = 1\) ----(2) Now, we have given that the intercept on the positive y-axis equal to twice its intercept on the positive x-axis. So, b = 2a Put b = 2a in equation 2 \(\frac{2}{{\rm{a}}} + \frac{3}{{2{\rm{a}}}} = 1\) ⇒ 2 + 3/2 = a ⇒ a = 7/2 Now, b = 2a = 7 So, equation of line will be, \(\frac{{\rm{x}}}{{\frac{7}{2}}} + \frac{{\rm{y}}}{7} = 1\) ⇒ 2x + y = 7 India’s #1 Learning Platform Start Complete Exam Preparation
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