Elements of analytical geometry and the differential and integral calculus . the length of the perpendicular from this given pointto the given straight line? ^^5^ _t_. 789^ 2. The equation of a line is y=—5x—15, and the co-ordinatesof a given point are x=:4 and y=5. What is the length of the perpendicular from the given pointto the straight line? Ans. PROPOSITION^ YIII. To find the equation of a straight line which will Used theangle contained hy the inclination of two other straight yz=^ax-\-h (1) And y=^ax-\h (2) be the equations of two straight lines which intersect, and
Elements of analytical geometry and the differential and integral calculus . the length of the perpendicular from this given pointto the given straight line? ^^5^ _t_. 789^ 2. The equation of a line is y=—5x—15, and the co-ordinatesof a given point are x=:4 and y=5. What is the length of the perpendicular from the given pointto the straight line? Ans. PROPOSITION^ YIII. To find the equation of a straight line which will Used theangle contained hy the inclination of two other straight yz=^ax-\-h (1) And y=^ax-\h (2) be the equations of two straight lines which intersect, and theco-ordinates of the point of intersection are .,=-(^) «-:^ (Prop. VI.)\a—a / a —a 22 ANALYTICAL GEOMETRY. We now require a third line which shall pass through thesame point of intersection and form an angle with the axis ofX (the tangent of which may be represented by m) which willbisect the angle made by the inclination of the other two by (Prop. V.) the equation of the line sought must be y—y i{x-^x^) (3) in case we can find the value of Let PN represent the linecorresponding to equation (1)»PM the line whose equation is(2), and PR the line required. Now the position or inclina-tion of PN to AX depends en-tirely on the value of a, and theinclination of Pilf depends on a\and all are entirely independentof the position of the point P, Now RPN=RPX—NPX and MPR=MPX^RPX. Whence by the application of a well known equation in planetrigonometry, (Equation (29), p. 143, in Robinsons Geometry,)we have m—a tan. RPX=tan.{RPX^XPX): 1-^am And tan. MPR=tan.{MPX—RPX)=- l-\-am But by hypothesis these two angles i^PA^and MPR are to beequal to each other. Therefore m—a a!—m Whence 1 -}-awi 1 -\-am2 , 2fl—aa) ,a-\-a (4) This equation will give two values of m ; one will correspondto the line PR, the other will be its supplement. If the proper value of m be taken from this equation and putin (3); then (3) will be the equation required. MiHiiHIMMiiMiMM
Size: 1550px × 1611px
Photo credit: © The Reading Room / Alamy / Afripics
License: Licensed
Model Released: No
Keywords: ., bookauthorrobinson, bookcentury1800, bookdecade1850, bookyear1856
